  
  [1X16 [33X[0;0YCombinatorics[133X[101X
  
  [33X[0;0YThis  chapter  describes  functions  that deal with combinatorics. We mainly
  concentrate  on two areas. One is about [13Xselections[113X, that is the ways one can
  select  elements from a set. The other is about [13Xpartitions[113X, that is the ways
  one can partition a set into the union of pairwise disjoint subsets.[133X
  
  
  [1X16.1 [33X[0;0YCombinatorial Numbers[133X[101X
  
  [1X16.1-1 Factorial[101X
  
  [33X[1;0Y[29X[2XFactorial[102X( [3Xn[103X ) [32X function[133X
  
  [33X[0;0Yreturns  the [13Xfactorial[113X [22Xn![122X of the positive integer [3Xn[103X, which is defined as the
  product [22X1 ⋅ 2 ⋅ 3 ⋯ n[122X.[133X
  
  [33X[0;0Y[22Xn![122X  is  the  number  of  permutations  of a set of [22Xn[122X elements. [22X1 / n![122X is the
  coefficient  of  [22Xx^n[122X  in  the  formal series [22Xexp(x)[122X, which is the generating
  function for factorial.[133X
  
  [4X[32X  Example  [32X[104X
    [4X[25Xgap>[125X [27XList( [0..10], Factorial );[127X[104X
    [4X[28X[ 1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800 ][128X[104X
    [4X[25Xgap>[125X [27XFactorial( 30 );[127X[104X
    [4X[28X265252859812191058636308480000000[128X[104X
  [4X[32X[104X
  
  [33X[0;0Y[2XPermutationsList[102X ([14X16.2-12[114X) computes the set of all permutations of a list.[133X
  
  [1X16.1-2 Binomial[101X
  
  [33X[1;0Y[29X[2XBinomial[102X( [3Xn[103X, [3Xk[103X ) [32X function[133X
  
  [33X[0;0Yreturns  the  [13Xbinomial coefficient[113X [22X{n choose k}[122X of integers [3Xn[103X and [3Xk[103X. This is
  defined by the conditions [22X{n choose k} = 0[122X for [22Xk < 0[122X, [22X{0 choose k} = 0[122X for [22Xk
  ≠  0[122X, [22X{0 choose 0} = 1[122X and the relation [22X{n choose k} = {n-1 choose k} + {n-1
  choose k-1}[122X for all [22Xn[122X and [22Xk[122X.[133X
  
  [33X[0;0YThere are many ways of describing this function. For example, if [22Xn ≥ 0[122X and [22X0
  ≤  k  ≤  n[122X,  then [22X{n choose k} = n! / (k! (n-k)!)[122X and for [22Xn < 0[122X and [22Xk ≥ 0[122X we
  have [22X{n choose k} = (-1)^k {-n+k-1 choose k}[122X.[133X
  
  [33X[0;0YIf [22Xn ≥ 0[122X then [22X{n choose k}[122X is the number of subsets with [22Xk[122X elements of a set
  with  [22Xn[122X  elements.  Also,  [22X{n  choose  k}[122X  is  the coefficient of [22Xx^k[122X in the
  polynomial  [22X(x  +  1)^n[122X,  which is the generating function for [22X{n choose .}[122X,
  hence the name.[133X
  
  [4X[32X  Example  [32X[104X
    [4X[25Xgap>[125X [27X# Knuth calls this the trademark of Binomial:[127X[104X
    [4X[25Xgap>[125X [27XList( [0..4], k->Binomial( 4, k ) );[127X[104X
    [4X[28X[ 1, 4, 6, 4, 1 ][128X[104X
    [4X[25Xgap>[125X [27XList( [0..6], n->List( [0..6], k->Binomial( n, k ) ) );;[127X[104X
    [4X[25Xgap>[125X [27X# the lower triangle is called Pascal's triangle:[127X[104X
    [4X[25Xgap>[125X [27XPrintArray( last );[127X[104X
    [4X[28X[ [   1,   0,   0,   0,   0,   0,   0 ],[128X[104X
    [4X[28X  [   1,   1,   0,   0,   0,   0,   0 ],[128X[104X
    [4X[28X  [   1,   2,   1,   0,   0,   0,   0 ],[128X[104X
    [4X[28X  [   1,   3,   3,   1,   0,   0,   0 ],[128X[104X
    [4X[28X  [   1,   4,   6,   4,   1,   0,   0 ],[128X[104X
    [4X[28X  [   1,   5,  10,  10,   5,   1,   0 ],[128X[104X
    [4X[28X  [   1,   6,  15,  20,  15,   6,   1 ] ][128X[104X
    [4X[25Xgap>[125X [27XBinomial( 50, 10 );[127X[104X
    [4X[28X10272278170[128X[104X
  [4X[32X[104X
  
  [33X[0;0Y[2XNrCombinations[102X  ([14X16.2-3[114X)  is  the  generalization of [2XBinomial[102X for multisets.
  [2XCombinations[102X ([14X16.2-1[114X) computes the set of all combinations of a multiset.[133X
  
  [1X16.1-3 Bell[101X
  
  [33X[1;0Y[29X[2XBell[102X( [3Xn[103X ) [32X function[133X
  
  [33X[0;0Yreturns  the  [13XBell number[113X [22XB(n)[122X. The Bell numbers are defined by [22XB(0) = 1[122X and
  the recurrence [22XB(n+1) = ∑_{k = 0}^n {n choose k} B(k)[122X.[133X
  
  [33X[0;0Y[22XB(n)[122X  is  the  number of ways to partition a set of [3Xn[103X elements into pairwise
  disjoint  nonempty  subsets  (see  [2XPartitionsSet[102X ([14X16.2-16[114X)). This implies of
  course that [22XB(n) = ∑_{k = 0}^n S_2(n,k)[122X (see [2XStirling2[102X ([14X16.1-6[114X)). [22XB(n)/n![122X is
  the  coefficient  of  [22Xx^n[122X in the formal series [22Xexp( exp(x)-1 )[122X, which is the
  generating function for [22XB(n)[122X.[133X
  
  [4X[32X  Example  [32X[104X
    [4X[25Xgap>[125X [27XList( [0..6], n -> Bell( n ) );[127X[104X
    [4X[28X[ 1, 1, 2, 5, 15, 52, 203 ][128X[104X
    [4X[25Xgap>[125X [27XBell( 14 );[127X[104X
    [4X[28X190899322[128X[104X
  [4X[32X[104X
  
  [1X16.1-4 Bernoulli[101X
  
  [33X[1;0Y[29X[2XBernoulli[102X( [3Xn[103X ) [32X function[133X
  
  [33X[0;0Yreturns the [3Xn[103X-th [13XBernoulli number[113X [22XB_n[122X, which is defined by [22XB_0 = 1[122X and [22XB_n =
  -∑_{k = 0}^{n-1} {n+1 choose k} B_k/(n+1)[122X.[133X
  
  [33X[0;0Y[22XB_n  /  n![122X  is the coefficient of [22Xx^n[122X in the power series of [22Xx / (exp(x)-1)[122X.
  Except for [22XB_1 = -1/2[122X the Bernoulli numbers for odd indices are zero.[133X
  
  [4X[32X  Example  [32X[104X
    [4X[25Xgap>[125X [27XBernoulli( 4 );[127X[104X
    [4X[28X-1/30[128X[104X
    [4X[25Xgap>[125X [27XBernoulli( 10 );[127X[104X
    [4X[28X5/66[128X[104X
    [4X[25Xgap>[125X [27XBernoulli( 12 );  # there is no simple pattern in Bernoulli numbers[127X[104X
    [4X[28X-691/2730[128X[104X
    [4X[25Xgap>[125X [27XBernoulli( 50 );  # and they grow fairly fast[127X[104X
    [4X[28X495057205241079648212477525/66[128X[104X
  [4X[32X[104X
  
  [1X16.1-5 Stirling1[101X
  
  [33X[1;0Y[29X[2XStirling1[102X( [3Xn[103X, [3Xk[103X ) [32X function[133X
  
  [33X[0;0Yreturns the [13XStirling number of the first kind[113X [22XS_1(n,k)[122X of the integers [3Xn[103X and
  [3Xk[103X.  Stirling numbers of the first kind are defined by [22XS_1(0,0) = 1[122X, [22XS_1(n,0)
  = S_1(0,k) = 0[122X if [22Xn, k ne 0[122X and the recurrence [22XS_1(n,k) = (n-1) S_1(n-1,k) +
  S_1(n-1,k-1)[122X.[133X
  
  [33X[0;0Y[22XS_1(n,k)[122X  is  the number of permutations of [3Xn[103X points with [3Xk[103X cycles. Stirling
  numbers  of the first kind appear as coefficients in the series [22Xn! {x choose
  n}  = ∑_{k = 0}^n S_1(n,k) x^k[122X which is the generating function for Stirling
  numbers of the first kind. Note the similarity to [22Xx^n = ∑_{k = 0}^n S_2(n,k)
  k! {x choose k}[122X (see [2XStirling2[102X ([14X16.1-6[114X)). Also the definition of [22XS_1[122X implies
  [22XS_1(n,k) = S_2(-k,-n)[122X if [22Xn, k < 0[122X. There are many formulae relating Stirling
  numbers  of  the  first  kind  to  Stirling numbers of the second kind, Bell
  numbers, and Binomial coefficients.[133X
  
  [4X[32X  Example  [32X[104X
    [4X[25Xgap>[125X [27X# Knuth calls this the trademark of S_1:[127X[104X
    [4X[25Xgap>[125X [27XList( [0..4], k -> Stirling1( 4, k ) );[127X[104X
    [4X[28X[ 0, 6, 11, 6, 1 ][128X[104X
    [4X[25Xgap>[125X [27XList( [0..6], n->List( [0..6], k->Stirling1( n, k ) ) );;[127X[104X
    [4X[25Xgap>[125X [27X# note the similarity with Pascal's triangle for Binomial numbers[127X[104X
    [4X[25Xgap>[125X [27XPrintArray( last );[127X[104X
    [4X[28X[ [    1,    0,    0,    0,    0,    0,    0 ],[128X[104X
    [4X[28X  [    0,    1,    0,    0,    0,    0,    0 ],[128X[104X
    [4X[28X  [    0,    1,    1,    0,    0,    0,    0 ],[128X[104X
    [4X[28X  [    0,    2,    3,    1,    0,    0,    0 ],[128X[104X
    [4X[28X  [    0,    6,   11,    6,    1,    0,    0 ],[128X[104X
    [4X[28X  [    0,   24,   50,   35,   10,    1,    0 ],[128X[104X
    [4X[28X  [    0,  120,  274,  225,   85,   15,    1 ] ][128X[104X
    [4X[25Xgap>[125X [27XStirling1(50,10);[127X[104X
    [4X[28X101623020926367490059043797119309944043405505380503665627365376[128X[104X
  [4X[32X[104X
  
  [1X16.1-6 Stirling2[101X
  
  [33X[1;0Y[29X[2XStirling2[102X( [3Xn[103X, [3Xk[103X ) [32X function[133X
  
  [33X[0;0Yreturns  the  [13XStirling  number of the second kind[113X [22XS_2(n,k)[122X of the integers [3Xn[103X
  and  [3Xk[103X.  Stirling  numbers  of  the second kind are defined by [22XS_2(0,0) = 1[122X,
  [22XS_2(n,0)  =  S_2(0,k)  =  0[122X  if  [22Xn,  k  ne 0[122X and the recurrence [22XS_2(n,k) = k
  S_2(n-1,k) + S_2(n-1,k-1)[122X.[133X
  
  [33X[0;0Y[22XS_2(n,k)[122X  is  the  number  of  ways  to partition a set of [3Xn[103X elements into [3Xk[103X
  pairwise  disjoint  nonempty subsets (see [2XPartitionsSet[102X ([14X16.2-16[114X)). Stirling
  numbers  of the second kind appear as coefficients in the expansion of [22Xx^n =
  ∑_{k = 0}^n S_2(n,k) k! {x choose k}[122X. Note the similarity to [22Xn! {x choose n}
  =  ∑_{k = 0}^n S_1(n,k) x^k[122X (see [2XStirling1[102X ([14X16.1-5[114X)). Also the definition of
  [22XS_2[122X  implies  [22XS_2(n,k)  =  S_1(-k,-n)[122X  if  [22Xn, k < 0[122X. There are many formulae
  relating  Stirling  numbers  of  the  second kind to Stirling numbers of the
  first kind, Bell numbers, and Binomial coefficients.[133X
  
  [4X[32X  Example  [32X[104X
    [4X[25Xgap>[125X [27X# Knuth calls this the trademark of S_2:[127X[104X
    [4X[25Xgap>[125X [27XList( [0..4], k->Stirling2( 4, k ) );[127X[104X
    [4X[28X[ 0, 1, 7, 6, 1 ][128X[104X
    [4X[25Xgap>[125X [27XList( [0..6], n->List( [0..6], k->Stirling2( n, k ) ) );;[127X[104X
    [4X[25Xgap>[125X [27X# note the similarity with Pascal's triangle for Binomial numbers[127X[104X
    [4X[25Xgap>[125X [27XPrintArray( last );[127X[104X
    [4X[28X[ [   1,   0,   0,   0,   0,   0,   0 ],[128X[104X
    [4X[28X  [   0,   1,   0,   0,   0,   0,   0 ],[128X[104X
    [4X[28X  [   0,   1,   1,   0,   0,   0,   0 ],[128X[104X
    [4X[28X  [   0,   1,   3,   1,   0,   0,   0 ],[128X[104X
    [4X[28X  [   0,   1,   7,   6,   1,   0,   0 ],[128X[104X
    [4X[28X  [   0,   1,  15,  25,  10,   1,   0 ],[128X[104X
    [4X[28X  [   0,   1,  31,  90,  65,  15,   1 ] ][128X[104X
    [4X[25Xgap>[125X [27XStirling2( 50, 10 );[127X[104X
    [4X[28X26154716515862881292012777396577993781727011[128X[104X
  [4X[32X[104X
  
  
  [1X16.2 [33X[0;0YCombinations, Arrangements and Tuples[133X[101X
  
  [1X16.2-1 Combinations[101X
  
  [33X[1;0Y[29X[2XCombinations[102X( [3Xmset[103X[, [3Xk[103X] ) [32X function[133X
  
  [33X[0;0Yreturns  the set of all combinations of the multiset [3Xmset[103X (a list of objects
  which  may  contain  the same object several times) with [3Xk[103X elements; if [3Xk[103X is
  not given it returns all combinations of [3Xmset[103X.[133X
  
  [33X[0;0YA  [13Xcombination[113X  of [3Xmset[103X is an unordered selection without repetitions and is
  represented  by a sorted sublist of [3Xmset[103X. If [3Xmset[103X is a proper set, there are
  [22X{|[3Xmset[103X|  choose [3Xk[103X}[122X (see [2XBinomial[102X ([14X16.1-2[114X)) combinations with [3Xk[103X elements, and
  the  set  of  all combinations is just the [13Xpower set[113X of [3Xmset[103X, which contains
  all [13Xsubsets[113X of [3Xmset[103X and has cardinality [22X2^{|[3Xmset[103X|}[122X.[133X
  
  [33X[0;0YTo  loop  over  combinations of a larger multiset use [2XIteratorOfCombinations[102X
  ([14X16.2-2[114X)  which  produces  combinations  one  by  one  and may save a lot of
  memory.   Another  memory  efficient  representation  of  the  list  of  all
  combinations is provided by [2XEnumeratorOfCombinations[102X ([14X16.2-2[114X).[133X
  
  
  [1X16.2-2 [33X[0;0YIterator and enumerator of combinations[133X[101X
  
  [33X[1;0Y[29X[2XIteratorOfCombinations[102X( [3Xmset[103X[, [3Xk[103X] ) [32X function[133X
  [33X[1;0Y[29X[2XEnumeratorOfCombinations[102X( [3Xmset[103X ) [32X function[133X
  
  [33X[0;0Y[2XIteratorOfCombinations[102X  returns  an  [2XIterator[102X ([14X30.8-1[114X) for combinations (see
  [2XCombinations[102X ([14X16.2-1[114X)) of the given multiset [3Xmset[103X. If a non-negative integer
  [3Xk[103X  is given as second argument then only the combinations with [3Xk[103X entries are
  produced, otherwise all combinations.[133X
  
  [33X[0;0Y[2XEnumeratorOfCombinations[102X   returns  an  [2XEnumerator[102X  ([14X30.3-2[114X)  of  the  given
  multiset  [3Xmset[103X.  Currently  only  a  variant  without  second  argument [3Xk[103X is
  implemented.[133X
  
  [33X[0;0YThe  ordering of combinations from these functions can be different and also
  different from the list returned by [2XCombinations[102X ([14X16.2-1[114X).[133X
  
  [4X[32X  Example  [32X[104X
    [4X[25Xgap>[125X [27Xm:=[1..15];; Add(m, 15);[127X[104X
    [4X[25Xgap>[125X [27XNrCombinations(m);[127X[104X
    [4X[28X49152[128X[104X
    [4X[25Xgap>[125X [27Xi := 0;; for c in Combinations(m) do i := i+1; od;[127X[104X
    [4X[25Xgap>[125X [27Xi;[127X[104X
    [4X[28X49152[128X[104X
    [4X[25Xgap>[125X [27Xcm := EnumeratorOfCombinations(m);;[127X[104X
    [4X[25Xgap>[125X [27Xcm[1000];[127X[104X
    [4X[28X[ 1, 2, 3, 6, 7, 8, 9, 10 ][128X[104X
    [4X[25Xgap>[125X [27XPosition(cm, [1,13,15,15]);[127X[104X
    [4X[28X36866[128X[104X
  [4X[32X[104X
  
  [1X16.2-3 NrCombinations[101X
  
  [33X[1;0Y[29X[2XNrCombinations[102X( [3Xmset[103X[, [3Xk[103X] ) [32X function[133X
  
  [33X[0;0Yreturns the number of [10XCombinations([3Xmset[103X[10X,[3Xk[103X[10X)[110X.[133X
  
  [4X[32X  Example  [32X[104X
    [4X[25Xgap>[125X [27XCombinations( [1,2,2,3] );[127X[104X
    [4X[28X[ [  ], [ 1 ], [ 1, 2 ], [ 1, 2, 2 ], [ 1, 2, 2, 3 ], [ 1, 2, 3 ], [128X[104X
    [4X[28X  [ 1, 3 ], [ 2 ], [ 2, 2 ], [ 2, 2, 3 ], [ 2, 3 ], [ 3 ] ][128X[104X
    [4X[25Xgap>[125X [27X# number of different hands in a game of poker:[127X[104X
    [4X[25Xgap>[125X [27XNrCombinations( [1..52], 5 );[127X[104X
    [4X[28X2598960[128X[104X
  [4X[32X[104X
  
  [33X[0;0YThe  function  [2XArrangements[102X  ([14X16.2-4[114X)  computes  ordered  selections without
  repetitions,  [2XUnorderedTuples[102X  ([14X16.2-6[114X)  computes  unordered selections with
  repetitions,   and   [2XTuples[102X   ([14X16.2-8[114X)   computes  ordered  selections  with
  repetitions.[133X
  
  [1X16.2-4 Arrangements[101X
  
  [33X[1;0Y[29X[2XArrangements[102X( [3Xmset[103X[, [3Xk[103X] ) [32X function[133X
  
  [33X[0;0Yreturns  the  set  of  arrangements  of  the  multiset  [3Xmset[103X  that contain [3Xk[103X
  elements. If [3Xk[103X is not given it returns all arrangements of [3Xmset[103X.[133X
  
  [33X[0;0YAn  [13Xarrangement[113X  of  [3Xmset[103X is an ordered selection without repetitions and is
  represented  by a list that contains only elements from [3Xmset[103X, but maybe in a
  different  order.  If  [3Xmset[103X  is a proper set there are [22X|mset|! / (|mset|-k)![122X
  (see [2XFactorial[102X ([14X16.1-1[114X)) arrangements with [3Xk[103X elements.[133X
  
  [1X16.2-5 NrArrangements[101X
  
  [33X[1;0Y[29X[2XNrArrangements[102X( [3Xmset[103X[, [3Xk[103X] ) [32X function[133X
  
  [33X[0;0Yreturns the number of [10XArrangements([3Xmset[103X[10X,[3Xk[103X[10X)[110X.[133X
  
  [33X[0;0YAs  an  example  of  arrangements of a multiset, think of the game Scrabble.
  Suppose  you  have  the  six characters of the word [10X"settle"[110X and you have to
  make a four letter word. Then the possibilities are given by[133X
  
  [4X[32X  Example  [32X[104X
    [4X[25Xgap>[125X [27XArrangements( ["s","e","t","t","l","e"], 4 );[127X[104X
    [4X[28X[ [ "e", "e", "l", "s" ], [ "e", "e", "l", "t" ], [ "e", "e", "s", "l" ],[128X[104X
    [4X[28X  [ "e", "e", "s", "t" ], [ "e", "e", "t", "l" ], [ "e", "e", "t", "s" ],[128X[104X
    [4X[28X  ... 93 more possibilities ...[128X[104X
    [4X[28X  [ "t", "t", "l", "s" ], [ "t", "t", "s", "e" ], [ "t", "t", "s", "l" ] ][128X[104X
  [4X[32X[104X
  
  [33X[0;0YCan  you  find  the  five proper English words, where [10X"lets"[110X does not count?
  Note  that  the  fact  that  the list returned by [2XArrangements[102X ([14X16.2-4[114X) is a
  proper  set  means  in this example that the possibilities are listed in the
  same order as they appear in the dictionary.[133X
  
  [4X[32X  Example  [32X[104X
    [4X[25Xgap>[125X [27XNrArrangements( ["s","e","t","t","l","e"] );[127X[104X
    [4X[28X523[128X[104X
  [4X[32X[104X
  
  [33X[0;0YThe  function  [2XCombinations[102X  ([14X16.2-1[114X)  computes unordered selections without
  repetitions,  [2XUnorderedTuples[102X  ([14X16.2-6[114X)  computes  unordered selections with
  repetitions,   and   [2XTuples[102X   ([14X16.2-8[114X)   computes  ordered  selections  with
  repetitions.[133X
  
  [1X16.2-6 UnorderedTuples[101X
  
  [33X[1;0Y[29X[2XUnorderedTuples[102X( [3Xset[103X, [3Xk[103X ) [32X function[133X
  
  [33X[0;0Yreturns the set of all unordered tuples of length [3Xk[103X of the set [3Xset[103X.[133X
  
  [33X[0;0YAn  [13Xunordered  tuple[113X  of  length  [3Xk[103X  of  [3Xset[103X  is an unordered selection with
  repetitions  of  [3Xset[103X  and  is  represented  by  a  sorted  list  of length [3Xk[103X
  containing  elements  from  [3Xset[103X.  There  are  [22X{|set|  + k - 1 choose k}[122X (see
  [2XBinomial[102X ([14X16.1-2[114X)) such unordered tuples.[133X
  
  [33X[0;0YNote  that the fact that [2XUnorderedTuples[102X returns a set implies that the last
  index runs fastest. That means the first tuple contains the smallest element
  from  [3Xset[103X  [3Xk[103X times, the second tuple contains the smallest element of [3Xset[103X at
  all  positions  except  at  the last positions, where it contains the second
  smallest element from [3Xset[103X and so on.[133X
  
  [1X16.2-7 NrUnorderedTuples[101X
  
  [33X[1;0Y[29X[2XNrUnorderedTuples[102X( [3Xset[103X, [3Xk[103X ) [32X function[133X
  
  [33X[0;0Yreturns the number of [10XUnorderedTuples([3Xset[103X[10X,[3Xk[103X[10X)[110X.[133X
  
  [33X[0;0YAs  an example for unordered tuples think of a poker-like game played with 5
  dice.  Then  each  possible hand corresponds to an unordered five-tuple from
  the set [22X{ 1, 2, ..., 6 }[122X.[133X
  
  [4X[32X  Example  [32X[104X
    [4X[25Xgap>[125X [27XNrUnorderedTuples( [1..6], 5 );[127X[104X
    [4X[28X252[128X[104X
    [4X[25Xgap>[125X [27XUnorderedTuples( [1..6], 5 );[127X[104X
    [4X[28X[ [ 1, 1, 1, 1, 1 ], [ 1, 1, 1, 1, 2 ], [ 1, 1, 1, 1, 3 ], [ 1, 1, 1, 1, 4 ],[128X[104X
    [4X[28X  [ 1, 1, 1, 1, 5 ], [ 1, 1, 1, 1, 6 ], [ 1, 1, 1, 2, 2 ], [ 1, 1, 1, 2, 3 ],[128X[104X
    [4X[28X  ... 100 more tuples ...[128X[104X
    [4X[28X  [ 1, 3, 5, 5, 6 ], [ 1, 3, 5, 6, 6 ], [ 1, 3, 6, 6, 6 ], [ 1, 4, 4, 4, 4 ],[128X[104X
    [4X[28X  ... 100 more tuples ...[128X[104X
    [4X[28X  [ 3, 3, 5, 5, 5 ], [ 3, 3, 5, 5, 6 ], [ 3, 3, 5, 6, 6 ], [ 3, 3, 6, 6, 6 ],[128X[104X
    [4X[28X  ... 32 more tuples ...[128X[104X
    [4X[28X  [ 5, 5, 5, 6, 6 ], [ 5, 5, 6, 6, 6 ], [ 5, 6, 6, 6, 6 ], [ 6, 6, 6, 6, 6 ] ][128X[104X
  [4X[32X[104X
  
  [33X[0;0YThe  function  [2XCombinations[102X  ([14X16.2-1[114X)  computes unordered selections without
  repetitions,  [2XArrangements[102X  ([14X16.2-4[114X)  computes  ordered  selections  without
  repetitions,   and   [2XTuples[102X   ([14X16.2-8[114X)   computes  ordered  selections  with
  repetitions.[133X
  
  [1X16.2-8 Tuples[101X
  
  [33X[1;0Y[29X[2XTuples[102X( [3Xset[103X, [3Xk[103X ) [32X function[133X
  
  [33X[0;0Yreturns the set of all ordered tuples of length [3Xk[103X of the set [3Xset[103X.[133X
  
  [33X[0;0YAn  [13Xordered tuple[113X of length [3Xk[103X of [3Xset[103X is an ordered selection with repetition
  and  is  represented by a list of length [3Xk[103X containing elements of [3Xset[103X. There
  are [22X|[3Xset[103X|^[3Xk[103X[122X such ordered tuples.[133X
  
  [33X[0;0YNote  that  the  fact  that [2XTuples[102X returns a set implies that the last index
  runs  fastest. That means the first tuple contains the smallest element from
  [3Xset[103X  [3Xk[103X  times,  the second tuple contains the smallest element of [3Xset[103X at all
  positions  except  at  the  last  positions,  where  it  contains the second
  smallest element from [3Xset[103X and so on.[133X
  
  [1X16.2-9 EnumeratorOfTuples[101X
  
  [33X[1;0Y[29X[2XEnumeratorOfTuples[102X( [3Xset[103X, [3Xk[103X ) [32X function[133X
  
  [33X[0;0YThis  function  is referred to as an example of enumerators that are defined
  by  functions  but are not constructed from a domain. The result is equal to
  that  of [10XTuples( [3Xset[103X[10X, [3Xk[103X[10X )[110X. However, the entries are not stored physically in
  the list but are created/identified on demand.[133X
  
  [1X16.2-10 IteratorOfTuples[101X
  
  [33X[1;0Y[29X[2XIteratorOfTuples[102X( [3Xset[103X, [3Xk[103X ) [32X function[133X
  
  [33X[0;0YFor a set [3Xset[103X and a positive integer [3Xk[103X, [2XIteratorOfTuples[102X returns an iterator
  (see [14X30.8[114X)  of the set of all ordered tuples (see [2XTuples[102X ([14X16.2-8[114X)) of length
  [3Xk[103X of the set [3Xset[103X. The tuples are returned in lexicographic order.[133X
  
  [1X16.2-11 NrTuples[101X
  
  [33X[1;0Y[29X[2XNrTuples[102X( [3Xset[103X, [3Xk[103X ) [32X function[133X
  
  [33X[0;0Yreturns the number of [10XTuples([3Xset[103X[10X,[3Xk[103X[10X)[110X.[133X
  
  [4X[32X  Example  [32X[104X
    [4X[25Xgap>[125X [27XTuples( [1,2,3], 2 );[127X[104X
    [4X[28X[ [ 1, 1 ], [ 1, 2 ], [ 1, 3 ], [ 2, 1 ], [ 2, 2 ], [ 2, 3 ], [128X[104X
    [4X[28X  [ 3, 1 ], [ 3, 2 ], [ 3, 3 ] ][128X[104X
    [4X[25Xgap>[125X [27XNrTuples( [1..10], 5 );[127X[104X
    [4X[28X100000[128X[104X
  [4X[32X[104X
  
  [33X[0;0Y[10XTuples([3Xset[103X[10X,[3Xk[103X[10X)[110X can also be viewed as the [3Xk[103X-fold cartesian product of [3Xset[103X (see
  [2XCartesian[102X ([14X21.20-16[114X)).[133X
  
  [33X[0;0YThe  function  [2XCombinations[102X  ([14X16.2-1[114X)  computes unordered selections without
  repetitions,  [2XArrangements[102X  ([14X16.2-4[114X)  computes  ordered  selections  without
  repetitions,  and  finally  the  function  [2XUnorderedTuples[102X ([14X16.2-6[114X) computes
  unordered selections with repetitions.[133X
  
  [1X16.2-12 PermutationsList[101X
  
  [33X[1;0Y[29X[2XPermutationsList[102X( [3Xmset[103X ) [32X function[133X
  
  [33X[0;0Y[2XPermutationsList[102X returns the set of permutations of the multiset [3Xmset[103X.[133X
  
  [33X[0;0YA  [13Xpermutation[113X  is  represented  by  a  list  that contains exactly the same
  elements  as  [3Xmset[103X, but possibly in different order. If [3Xmset[103X is a proper set
  there  are [22X|[3Xmset[103X| ![122X (see [2XFactorial[102X ([14X16.1-1[114X)) such permutations. Otherwise if
  the  first  elements appears [22Xk_1[122X times, the second element appears [22Xk_2[122X times
  and  so  on, the number of permutations is [22X|[3Xmset[103X| ! / (k_1! k_2! ...)[122X, which
  is sometimes called multinomial coefficient.[133X
  
  [1X16.2-13 NrPermutationsList[101X
  
  [33X[1;0Y[29X[2XNrPermutationsList[102X( [3Xmset[103X ) [32X function[133X
  
  [33X[0;0Yreturns the number of [10XPermutationsList([3Xmset[103X[10X)[110X.[133X
  
  [4X[32X  Example  [32X[104X
    [4X[25Xgap>[125X [27XPermutationsList( [1,2,3] );[127X[104X
    [4X[28X[ [ 1, 2, 3 ], [ 1, 3, 2 ], [ 2, 1, 3 ], [ 2, 3, 1 ], [ 3, 1, 2 ], [128X[104X
    [4X[28X  [ 3, 2, 1 ] ][128X[104X
    [4X[25Xgap>[125X [27XPermutationsList( [1,1,2,2] );[127X[104X
    [4X[28X[ [ 1, 1, 2, 2 ], [ 1, 2, 1, 2 ], [ 1, 2, 2, 1 ], [ 2, 1, 1, 2 ], [128X[104X
    [4X[28X  [ 2, 1, 2, 1 ], [ 2, 2, 1, 1 ] ][128X[104X
    [4X[25Xgap>[125X [27XNrPermutationsList( [1,2,2,3,3,3,4,4,4,4] );[127X[104X
    [4X[28X12600[128X[104X
  [4X[32X[104X
  
  [33X[0;0YThe function [2XArrangements[102X ([14X16.2-4[114X) is the generalization of [2XPermutationsList[102X
  ([14X16.2-12[114X)  that  allows  you  to  specify  the  size  of  the  permutations.
  [2XDerangements[102X ([14X16.2-14[114X) computes permutations that have no fixed points.[133X
  
  [1X16.2-14 Derangements[101X
  
  [33X[1;0Y[29X[2XDerangements[102X( [3Xlist[103X ) [32X function[133X
  
  [33X[0;0Yreturns the set of all derangements of the list [3Xlist[103X.[133X
  
  [33X[0;0YA  [13Xderangement[113X is a fixpointfree permutation of [3Xlist[103X and is represented by a
  list  that  contains exactly the same elements as [3Xlist[103X, but in such an order
  that  the  derangement  has  at no position the same element as [3Xlist[103X. If the
  list [3Xlist[103X contains no element twice there are exactly [22X|[3Xlist[103X|! (1/2! - 1/3! +
  1/4! - ⋯ + (-1)^n / n!)[122X derangements.[133X
  
  [33X[0;0YNote  that  the ratio [10XNrPermutationsList( [ 1 .. n ] ) / NrDerangements( [ 1
  ..  n  ]  )[110X, which is [22Xn! / (n! (1/2! - 1/3! + 1/4! - ⋯ + (-1)^n / n!))[122X is an
  approximation  for  the  base  of the natural logarithm [22Xe = 2.7182818285...[122X,
  which is correct to about [22Xn[122X digits.[133X
  
  [1X16.2-15 NrDerangements[101X
  
  [33X[1;0Y[29X[2XNrDerangements[102X( [3Xlist[103X ) [32X function[133X
  
  [33X[0;0Yreturns the number of [10XDerangements([3Xlist[103X[10X)[110X.[133X
  
  [33X[0;0YAs  an  example of derangements suppose that you have to send four different
  letters to four different people. Then a derangement corresponds to a way to
  send those letters such that no letter reaches the intended person.[133X
  
  [4X[32X  Example  [32X[104X
    [4X[25Xgap>[125X [27XDerangements( [1,2,3,4] );[127X[104X
    [4X[28X[ [ 2, 1, 4, 3 ], [ 2, 3, 4, 1 ], [ 2, 4, 1, 3 ], [ 3, 1, 4, 2 ], [128X[104X
    [4X[28X  [ 3, 4, 1, 2 ], [ 3, 4, 2, 1 ], [ 4, 1, 2, 3 ], [ 4, 3, 1, 2 ], [128X[104X
    [4X[28X  [ 4, 3, 2, 1 ] ][128X[104X
    [4X[25Xgap>[125X [27XNrDerangements( [1..10] );[127X[104X
    [4X[28X1334961[128X[104X
    [4X[25Xgap>[125X [27XInt( 10^7*NrPermutationsList([1..10])/last );[127X[104X
    [4X[28X27182816[128X[104X
    [4X[25Xgap>[125X [27XDerangements( [1,1,2,2,3,3] );[127X[104X
    [4X[28X[ [ 2, 2, 3, 3, 1, 1 ], [ 2, 3, 1, 3, 1, 2 ], [ 2, 3, 1, 3, 2, 1 ], [128X[104X
    [4X[28X  [ 2, 3, 3, 1, 1, 2 ], [ 2, 3, 3, 1, 2, 1 ], [ 3, 2, 1, 3, 1, 2 ], [128X[104X
    [4X[28X  [ 3, 2, 1, 3, 2, 1 ], [ 3, 2, 3, 1, 1, 2 ], [ 3, 2, 3, 1, 2, 1 ], [128X[104X
    [4X[28X  [ 3, 3, 1, 1, 2, 2 ] ][128X[104X
    [4X[25Xgap>[125X [27XNrDerangements( [1,2,2,3,3,3,4,4,4,4] );[127X[104X
    [4X[28X338[128X[104X
  [4X[32X[104X
  
  [33X[0;0YThe function [2XPermutationsList[102X ([14X16.2-12[114X) computes all permutations of a list.[133X
  
  [1X16.2-16 PartitionsSet[101X
  
  [33X[1;0Y[29X[2XPartitionsSet[102X( [3Xset[103X[, [3Xk[103X] ) [32X function[133X
  
  [33X[0;0Yreturns  the  set of all unordered partitions of the set [3Xset[103X into [3Xk[103X pairwise
  disjoint  nonempty  sets.  If  [3Xk[103X  is  not  given  it  returns  all unordered
  partitions of [3Xset[103X for all [3Xk[103X.[133X
  
  [33X[0;0YAn  [13Xunordered  partition[113X  of [3Xset[103X is a set of pairwise disjoint nonempty sets
  with  union  [3Xset[103X and is represented by a sorted list of such sets. There are
  [22XB( |set| )[122X (see [2XBell[102X ([14X16.1-3[114X)) partitions of the set [3Xset[103X and [22XS_2( |set|, k )[122X
  (see [2XStirling2[102X ([14X16.1-6[114X)) partitions with [3Xk[103X elements.[133X
  
  [1X16.2-17 NrPartitionsSet[101X
  
  [33X[1;0Y[29X[2XNrPartitionsSet[102X( [3Xset[103X[, [3Xk[103X] ) [32X function[133X
  
  [33X[0;0Yreturns the number of [10XPartitionsSet([3Xset[103X[10X,[3Xk[103X[10X)[110X.[133X
  
  [4X[32X  Example  [32X[104X
    [4X[25Xgap>[125X [27XPartitionsSet( [1,2,3] );[127X[104X
    [4X[28X[ [ [ 1 ], [ 2 ], [ 3 ] ], [ [ 1 ], [ 2, 3 ] ], [ [ 1, 2 ], [ 3 ] ], [128X[104X
    [4X[28X  [ [ 1, 2, 3 ] ], [ [ 1, 3 ], [ 2 ] ] ][128X[104X
    [4X[25Xgap>[125X [27XPartitionsSet( [1,2,3,4], 2 );[127X[104X
    [4X[28X[ [ [ 1 ], [ 2, 3, 4 ] ], [ [ 1, 2 ], [ 3, 4 ] ], [128X[104X
    [4X[28X  [ [ 1, 2, 3 ], [ 4 ] ], [ [ 1, 2, 4 ], [ 3 ] ], [128X[104X
    [4X[28X  [ [ 1, 3 ], [ 2, 4 ] ], [ [ 1, 3, 4 ], [ 2 ] ], [128X[104X
    [4X[28X  [ [ 1, 4 ], [ 2, 3 ] ] ][128X[104X
    [4X[25Xgap>[125X [27XNrPartitionsSet( [1..6] );[127X[104X
    [4X[28X203[128X[104X
    [4X[25Xgap>[125X [27XNrPartitionsSet( [1..10], 3 );[127X[104X
    [4X[28X9330[128X[104X
  [4X[32X[104X
  
  [33X[0;0YNote  that  [2XPartitionsSet[102X ([14X16.2-16[114X) does currently not support multisets and
  that there is currently no ordered counterpart.[133X
  
  [1X16.2-18 Partitions[101X
  
  [33X[1;0Y[29X[2XPartitions[102X( [3Xn[103X[, [3Xk[103X] ) [32X function[133X
  
  [33X[0;0Yreturns the set of all (unordered) partitions of the positive integer [3Xn[103X into
  sums  with [3Xk[103X summands. If [3Xk[103X is not given it returns all unordered partitions
  of [3Xset[103X for all [3Xk[103X.[133X
  
  [33X[0;0YAn  [13Xunordered  partition[113X  is  an  unordered  sum  [22Xn = p_1 + p_2 + ⋯ + p_k[122X of
  positive integers and is represented by the list [22Xp = [ p_1, p_2, ..., p_k ][122X,
  in  nonincreasing  order, i.e., [22Xp_1 ≥ p_2 ≥ ... ≥ p_k[122X. We write [22Xp ⊢ n[122X. There
  are  approximately  [22Xexp(π  sqrt{2/3 n}) / (4 sqrt{3} n)[122X such partitions, use
  [2XNrPartitions[102X ([14X16.2-20[114X) to compute the precise number.[133X
  
  [33X[0;0YIf you want to loop over all partitions of some larger [3Xn[103X use the more memory
  efficient [2XIteratorOfPartitions[102X ([14X16.2-19[114X).[133X
  
  [33X[0;0YIt  is  possible  to  associate  with  every  partition  of  the integer [3Xn[103X a
  conjugacy  class of permutations in the symmetric group on [3Xn[103X points and vice
  versa.  Therefore  [22Xp(n) :=[122X[10XNrPartitions[110X[22X(n)[122X is the number of conjugacy classes
  of the symmetric group on [3Xn[103X points.[133X
  
  [33X[0;0YRamanujan  found  the  identities  [22Xp(5i+4)  = 0[122X mod 5, [22Xp(7i+5) = 0[122X mod 7 and
  [22Xp(11i+6)  =  0[122X  mod 11 and many other fascinating things about the number of
  partitions.[133X
  
  [1X16.2-19 IteratorOfPartitions[101X
  
  [33X[1;0Y[29X[2XIteratorOfPartitions[102X( [3Xn[103X ) [32X function[133X
  
  [33X[0;0YFor   a   positive  integer  [3Xn[103X,  [2XIteratorOfPartitions[102X  returns  an  iterator
  (see [14X30.8[114X)  of  the  set  of partitions of [3Xn[103X (see [2XPartitions[102X ([14X16.2-18[114X)). The
  partitions of [3Xn[103X are returned in lexicographic order.[133X
  
  [1X16.2-20 NrPartitions[101X
  
  [33X[1;0Y[29X[2XNrPartitions[102X( [3Xn[103X[, [3Xk[103X] ) [32X function[133X
  
  [33X[0;0Yreturns the number of [10XPartitions([3Xset[103X[10X,[3Xk[103X[10X)[110X.[133X
  
  [4X[32X  Example  [32X[104X
    [4X[25Xgap>[125X [27XPartitions( 7 );[127X[104X
    [4X[28X[ [ 1, 1, 1, 1, 1, 1, 1 ], [ 2, 1, 1, 1, 1, 1 ], [ 2, 2, 1, 1, 1 ], [128X[104X
    [4X[28X  [ 2, 2, 2, 1 ], [ 3, 1, 1, 1, 1 ], [ 3, 2, 1, 1 ], [ 3, 2, 2 ], [128X[104X
    [4X[28X  [ 3, 3, 1 ], [ 4, 1, 1, 1 ], [ 4, 2, 1 ], [ 4, 3 ], [ 5, 1, 1 ], [128X[104X
    [4X[28X  [ 5, 2 ], [ 6, 1 ], [ 7 ] ][128X[104X
    [4X[25Xgap>[125X [27XPartitions( 8, 3 );[127X[104X
    [4X[28X[ [ 3, 3, 2 ], [ 4, 2, 2 ], [ 4, 3, 1 ], [ 5, 2, 1 ], [ 6, 1, 1 ] ][128X[104X
    [4X[25Xgap>[125X [27XNrPartitions( 7 );[127X[104X
    [4X[28X15[128X[104X
    [4X[25Xgap>[125X [27XNrPartitions( 100 );[127X[104X
    [4X[28X190569292[128X[104X
  [4X[32X[104X
  
  [33X[0;0YThe  function  [2XOrderedPartitions[102X  ([14X16.2-21[114X)  is  the  ordered counterpart of
  [2XPartitions[102X ([14X16.2-18[114X).[133X
  
  [1X16.2-21 OrderedPartitions[101X
  
  [33X[1;0Y[29X[2XOrderedPartitions[102X( [3Xn[103X[, [3Xk[103X] ) [32X function[133X
  
  [33X[0;0Yreturns  the  set  of  all ordered partitions of the positive integer [3Xn[103X into
  sums with [3Xk[103X summands. If [3Xk[103X is not given it returns all ordered partitions of
  [3Xset[103X for all [3Xk[103X.[133X
  
  [33X[0;0YAn [13Xordered partition[113X is an ordered sum [22Xn = p_1 + p_2 + ... + p_k[122X of positive
  integers  and  is  represented by the list [22X[ p_1, p_2, ..., p_k ][122X. There are
  totally  [22X2^{n-1}[122X  ordered  partitions  and  [22X{n-1  choose  k-1}[122X (see [2XBinomial[102X
  ([14X16.1-2[114X)) ordered partitions with [3Xk[103X summands.[133X
  
  [33X[0;0YDo  not  call [2XOrderedPartitions[102X with an [3Xn[103X much larger than [22X15[122X, the list will
  simply become too large.[133X
  
  [1X16.2-22 NrOrderedPartitions[101X
  
  [33X[1;0Y[29X[2XNrOrderedPartitions[102X( [3Xn[103X[, [3Xk[103X] ) [32X function[133X
  
  [33X[0;0Yreturns the number of [10XOrderedPartitions([3Xset[103X[10X,[3Xk[103X[10X)[110X.[133X
  
  [4X[32X  Example  [32X[104X
    [4X[25Xgap>[125X [27XOrderedPartitions( 5 );[127X[104X
    [4X[28X[ [ 1, 1, 1, 1, 1 ], [ 1, 1, 1, 2 ], [ 1, 1, 2, 1 ], [ 1, 1, 3 ], [128X[104X
    [4X[28X  [ 1, 2, 1, 1 ], [ 1, 2, 2 ], [ 1, 3, 1 ], [ 1, 4 ], [ 2, 1, 1, 1 ], [128X[104X
    [4X[28X  [ 2, 1, 2 ], [ 2, 2, 1 ], [ 2, 3 ], [ 3, 1, 1 ], [ 3, 2 ], [128X[104X
    [4X[28X  [ 4, 1 ], [ 5 ] ][128X[104X
    [4X[25Xgap>[125X [27XOrderedPartitions( 6, 3 );[127X[104X
    [4X[28X[ [ 1, 1, 4 ], [ 1, 2, 3 ], [ 1, 3, 2 ], [ 1, 4, 1 ], [ 2, 1, 3 ], [128X[104X
    [4X[28X  [ 2, 2, 2 ], [ 2, 3, 1 ], [ 3, 1, 2 ], [ 3, 2, 1 ], [ 4, 1, 1 ] ][128X[104X
    [4X[25Xgap>[125X [27XNrOrderedPartitions(20);[127X[104X
    [4X[28X524288[128X[104X
  [4X[32X[104X
  
  [33X[0;0YThe   function   [2XPartitions[102X   ([14X16.2-18[114X)  is  the  unordered  counterpart  of
  [2XOrderedPartitions[102X ([14X16.2-21[114X).[133X
  
  [1X16.2-23 PartitionsGreatestLE[101X
  
  [33X[1;0Y[29X[2XPartitionsGreatestLE[102X( [3Xn[103X, [3Xm[103X ) [32X function[133X
  
  [33X[0;0Yreturns  the set of all (unordered) partitions of the integer [3Xn[103X having parts
  less or equal to the integer [3Xm[103X.[133X
  
  [1X16.2-24 PartitionsGreatestEQ[101X
  
  [33X[1;0Y[29X[2XPartitionsGreatestEQ[102X( [3Xn[103X, [3Xm[103X ) [32X function[133X
  
  [33X[0;0Yreturns  the  set  of  all  (unordered)  partitions  of the integer [3Xn[103X having
  greatest part equal to the integer [3Xm[103X.[133X
  
  [1X16.2-25 RestrictedPartitions[101X
  
  [33X[1;0Y[29X[2XRestrictedPartitions[102X( [3Xn[103X, [3Xset[103X[, [3Xk[103X] ) [32X function[133X
  
  [33X[0;0YIn  the  first  form  [2XRestrictedPartitions[102X returns the set of all restricted
  partitions  of  the  positive  integer  [3Xn[103X into sums with [3Xk[103X summands with the
  summands  of  the  partition  coming from the set [3Xset[103X. If [3Xk[103X is not given all
  restricted partitions for all [3Xk[103X are returned.[133X
  
  [33X[0;0YA  [13Xrestricted  partition[113X  is  like  an  ordinary  partition  (see [2XPartitions[102X
  ([14X16.2-18[114X))  an  unordered sum [22Xn = p_1 + p_2 + ... + p_k[122X of positive integers
  and  is represented by the list [22Xp = [ p_1, p_2, ..., p_k ][122X, in nonincreasing
  order.  The  difference  is  that here the [22Xp_i[122X must be elements from the set
  [3Xset[103X, while for ordinary partitions they may be elements from [10X[ 1 .. n ][110X.[133X
  
  [1X16.2-26 NrRestrictedPartitions[101X
  
  [33X[1;0Y[29X[2XNrRestrictedPartitions[102X( [3Xn[103X, [3Xset[103X[, [3Xk[103X] ) [32X function[133X
  
  [33X[0;0Yreturns the number of [10XRestrictedPartitions([3Xn[103X[10X,[3Xset[103X[10X,[3Xk[103X[10X)[110X.[133X
  
  [4X[32X  Example  [32X[104X
    [4X[25Xgap>[125X [27XRestrictedPartitions( 8, [1,3,5,7] );[127X[104X
    [4X[28X[ [ 1, 1, 1, 1, 1, 1, 1, 1 ], [ 3, 1, 1, 1, 1, 1 ], [ 3, 3, 1, 1 ], [128X[104X
    [4X[28X  [ 5, 1, 1, 1 ], [ 5, 3 ], [ 7, 1 ] ][128X[104X
    [4X[25Xgap>[125X [27XNrRestrictedPartitions(50,[1,2,5,10,20,50]);[127X[104X
    [4X[28X451[128X[104X
  [4X[32X[104X
  
  [33X[0;0YThe  last example tells us that there are 451 ways to return 50 pence change
  using 1, 2, 5, 10, 20 and 50 pence coins.[133X
  
  [1X16.2-27 SignPartition[101X
  
  [33X[1;0Y[29X[2XSignPartition[102X( [3Xpi[103X ) [32X function[133X
  
  [33X[0;0Yreturns the sign of a permutation with cycle structure [3Xpi[103X.[133X
  
  [33X[0;0YThis function actually describes a homomorphism from the symmetric group [22XS_n[122X
  into  the  cyclic  group of order 2, whose kernel is exactly the alternating
  group  [22XA_n[122X  (see  [2XSignPerm[102X  ([14X42.4-1[114X)).  Partitions of sign 1 are called [13Xeven[113X
  partitions while partitions of sign [22X-1[122X are called [13Xodd[113X.[133X
  
  [4X[32X  Example  [32X[104X
    [4X[25Xgap>[125X [27XSignPartition([6,5,4,3,2,1]);[127X[104X
    [4X[28X-1[128X[104X
  [4X[32X[104X
  
  [1X16.2-28 AssociatedPartition[101X
  
  [33X[1;0Y[29X[2XAssociatedPartition[102X( [3Xpi[103X ) [32X function[133X
  
  [33X[0;0Y[2XAssociatedPartition[102X  returns  the  associated  partition of the partition [3Xpi[103X
  which is obtained by transposing the corresponding Young diagram.[133X
  
  [4X[32X  Example  [32X[104X
    [4X[25Xgap>[125X [27XAssociatedPartition([4,2,1]);[127X[104X
    [4X[28X[ 3, 2, 1, 1 ][128X[104X
    [4X[25Xgap>[125X [27XAssociatedPartition([6]);[127X[104X
    [4X[28X[ 1, 1, 1, 1, 1, 1 ][128X[104X
  [4X[32X[104X
  
  [1X16.2-29 PowerPartition[101X
  
  [33X[1;0Y[29X[2XPowerPartition[102X( [3Xpi[103X, [3Xk[103X ) [32X function[133X
  
  [33X[0;0Y[2XPowerPartition[102X  returns  the  partition corresponding to the [3Xk[103X-th power of a
  permutation with cycle structure [3Xpi[103X.[133X
  
  [33X[0;0YEach  part  [22Xl[122X  of  [3Xpi[103X  is replaced by [22Xd = gcd(l, k)[122X parts [22Xl/d[122X. So if [3Xpi[103X is a
  partition  of [22Xn[122X then [22X[3Xpi[103X^[3Xk[103X[122X also is a partition of [22Xn[122X. [2XPowerPartition[102X describes
  the power map of symmetric groups.[133X
  
  [4X[32X  Example  [32X[104X
    [4X[25Xgap>[125X [27XPowerPartition([6,5,4,3,2,1], 3);[127X[104X
    [4X[28X[ 5, 4, 2, 2, 2, 2, 1, 1, 1, 1 ][128X[104X
  [4X[32X[104X
  
  [1X16.2-30 PartitionTuples[101X
  
  [33X[1;0Y[29X[2XPartitionTuples[102X( [3Xn[103X, [3Xr[103X ) [32X function[133X
  
  [33X[0;0Y[2XPartitionTuples[102X  returns  the  list  of  all  [3Xr[103X-tuples  of  partitions which
  together form a partition of [3Xn[103X.[133X
  
  [33X[0;0Y[3Xr[103X-tuples  of  partitions  describe  the classes and the characters of wreath
  products  of  groups  with [3Xr[103X conjugacy classes with the symmetric group on [3Xn[103X
  points,       see      [2XCharacterTableWreathSymmetric[102X      ([14X71.20-6[114X)      and
  [2XCharacterValueWreathSymmetric[102X ([14X71.20-7[114X).[133X
  
  [1X16.2-31 NrPartitionTuples[101X
  
  [33X[1;0Y[29X[2XNrPartitionTuples[102X( [3Xn[103X, [3Xr[103X ) [32X function[133X
  
  [33X[0;0Yreturns the number of [10XPartitionTuples( [3Xn[103X[10X, [3Xr[103X[10X )[110X.[133X
  
  [4X[32X  Example  [32X[104X
    [4X[25Xgap>[125X [27XPartitionTuples(3, 2);[127X[104X
    [4X[28X[ [ [ 1, 1, 1 ], [  ] ], [ [ 1, 1 ], [ 1 ] ], [ [ 1 ], [ 1, 1 ] ], [128X[104X
    [4X[28X  [ [  ], [ 1, 1, 1 ] ], [ [ 2, 1 ], [  ] ], [ [ 1 ], [ 2 ] ], [128X[104X
    [4X[28X  [ [ 2 ], [ 1 ] ], [ [  ], [ 2, 1 ] ], [ [ 3 ], [  ] ], [128X[104X
    [4X[28X  [ [  ], [ 3 ] ] ][128X[104X
  [4X[32X[104X
  
  [1X16.2-32 BetaSet[101X
  
  [33X[1;0Y[29X[2XBetaSet[102X( [3Xalpha[103X ) [32X function[133X
  
  [33X[0;0YFor  a  list  [3Xalpha[103X that describes a partition of a nonnegative integer (see
  [2XPartitions[102X  ([14X16.2-18[114X)),  [2XBetaSet[102X  returns  the  list of integers obtained by
  reversing  the  order of [3Xalpha[103X and then adding the sequence [10X[ 0, 1, 2, ... ][110X
  of the same length, cf. [JK81, Section 2.7].[133X
  
  [4X[32X  Example  [32X[104X
    [4X[25Xgap>[125X [27XBetaSet( [ 4, 2, 1 ] );[127X[104X
    [4X[28X[ 1, 3, 6 ][128X[104X
    [4X[25Xgap>[125X [27XBetaSet( [] );[127X[104X
    [4X[28X[  ][128X[104X
  [4X[32X[104X
  
  
  [1X16.3 [33X[0;0YFibonacci and Lucas Sequences[133X[101X
  
  [1X16.3-1 Fibonacci[101X
  
  [33X[1;0Y[29X[2XFibonacci[102X( [3Xn[103X ) [32X function[133X
  
  [33X[0;0Yreturns the [3Xn[103Xth number of the [13XFibonacci sequence[113X. The Fibonacci sequence [22XF_n[122X
  is  defined  by  the  initial  conditions  [22XF_1  = F_2 = 1[122X and the recurrence
  relation  [22XF_{n+2} = F_{n+1} + F_n[122X. For negative [22Xn[122X we define [22XF_n = (-1)^{n+1}
  F_{-n}[122X, which is consistent with the recurrence relation.[133X
  
  [33X[0;0YUsing  generating functions one can prove that [22XF_n = ϕ^n - 1/ϕ^n[122X, where [22Xϕ[122X is
  [22X(sqrt{5}  +  1)/2[122X, i.e., one root of [22Xx^2 - x - 1 = 0[122X. Fibonacci numbers have
  the property [22Xgcd( F_m, F_n ) = F_{gcd(m,n)}[122X. But a pair of Fibonacci numbers
  requires  more  division steps in Euclid's algorithm (see [2XGcd[102X ([14X56.7-1[114X)) than
  any  other  pair  of  integers of the same size. [10XFibonacci([3Xk[103X[10X)[110X is the special
  case [10XLucas(1,-1,[3Xk[103X[10X)[1][110X (see [2XLucas[102X ([14X16.3-2[114X)).[133X
  
  [4X[32X  Example  [32X[104X
    [4X[25Xgap>[125X [27XFibonacci( 10 );[127X[104X
    [4X[28X55[128X[104X
    [4X[25Xgap>[125X [27XFibonacci( 35 );[127X[104X
    [4X[28X9227465[128X[104X
    [4X[25Xgap>[125X [27XFibonacci( -10 );[127X[104X
    [4X[28X-55[128X[104X
  [4X[32X[104X
  
  [1X16.3-2 Lucas[101X
  
  [33X[1;0Y[29X[2XLucas[102X( [3XP[103X, [3XQ[103X, [3Xk[103X ) [32X function[133X
  
  [33X[0;0Yreturns the [3Xk[103X-th values of the [13XLucas sequence[113X with parameters [3XP[103X and [3XQ[103X, which
  must  be  integers, as a list of three integers. If [3Xk[103X is a negative integer,
  then  the  values of the Lucas sequence may be nonintegral rational numbers,
  with denominator roughly [3XQ[103X^[3Xk[103X.[133X
  
  [33X[0;0YLet  [22Xα,  β[122X  be  the two roots of [22Xx^2 - P x + Q[122X then we define [10XLucas( [3XP[103X[10X, [3XQ[103X[10X, [3Xk[103X[10X
  )[1][110X  [22X=  U_k  = (α^k - β^k) / (α - β)[122X and [10XLucas( [3XP[103X[10X, [3XQ[103X[10X, [3Xk[103X[10X )[2][110X [22X= V_k = (α^k +
  β^k)[122X and as a convenience [10XLucas( [3XP[103X[10X, [3XQ[103X[10X, [3Xk[103X[10X )[3][110X [22X= Q^k[122X.[133X
  
  [33X[0;0YThe  following  recurrence  relations are easily derived from the definition
  [22XU_0  = 0, U_1 = 1, U_k = P U_{k-1} - Q U_{k-2}[122X and [22XV_0 = 2, V_1 = P, V_k = P
  V_{k-1}  - Q V_{k-2}[122X. Those relations are actually used to define [2XLucas[102X if [22Xα
  = β[122X.[133X
  
  [33X[0;0YAlso  the  more complex relations used in [2XLucas[102X can be easily derived [22XU_2k =
  U_k V_k[122X, [22XU_{2k+1} = (P U_2k + V_2k) / 2[122X and [22XV_2k = V_k^2 - 2 Q^k[122X, [22XV_{2k+1} =
  ((P^2-4Q) U_2k + P V_2k) / 2[122X.[133X
  
  [33X[0;0Y[10XFibonacci([3Xk[103X[10X)[110X  (see  [2XFibonacci[102X  ([14X16.3-1[114X))  is  simply [10XLucas(1,-1,[3Xk[103X[10X)[1][110X. In an
  abuse  of  notation,  the  sequence [10XLucas(1,-1,[3Xk[103X[10X)[2][110X is sometimes called the
  Lucas sequence.[133X
  
  [4X[32X  Example  [32X[104X
    [4X[25Xgap>[125X [27XList( [0..10], i -> Lucas(1,-2,i)[1] );     # 2^k - (-1)^k)/3[127X[104X
    [4X[28X[ 0, 1, 1, 3, 5, 11, 21, 43, 85, 171, 341 ][128X[104X
    [4X[25Xgap>[125X [27XList( [0..10], i -> Lucas(1,-2,i)[2] );     # 2^k + (-1)^k[127X[104X
    [4X[28X[ 2, 1, 5, 7, 17, 31, 65, 127, 257, 511, 1025 ][128X[104X
    [4X[25Xgap>[125X [27XList( [0..10], i -> Lucas(1,-1,i)[1] );     # Fibonacci sequence[127X[104X
    [4X[28X[ 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55 ][128X[104X
    [4X[25Xgap>[125X [27XList( [0..10], i -> Lucas(2,1,i)[1] );      # the roots are equal[127X[104X
    [4X[28X[ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ][128X[104X
  [4X[32X[104X
  
  
  [1X16.4 [33X[0;0YPermanent of a Matrix[133X[101X
  
  [1X16.4-1 Permanent[101X
  
  [33X[1;0Y[29X[2XPermanent[102X( [3Xmat[103X ) [32X attribute[133X
  
  [33X[0;0Yreturns  the [13Xpermanent[113X of the matrix [3Xmat[103X. The permanent is defined by [22X∑_{p ∈
  Sym(n)} ∏_{i = 1}^n mat[i][i^p][122X.[133X
  
  [33X[0;0YNote  the similarity of the definition of the permanent to the definition of
  the  determinant  (see [2XDeterminantMat[102X ([14X24.4-4[114X)). In fact the only difference
  is  the  missing  sign  of  the  permutation. However the permanent is quite
  unlike the determinant, for example it is not multilinear or alternating. It
  has however important combinatorial properties.[133X
  
  [4X[32X  Example  [32X[104X
    [4X[25Xgap>[125X [27XPermanent( [[0,1,1,1],[127X[104X
    [4X[25X>[125X [27X     [1,0,1,1],[127X[104X
    [4X[25X>[125X [27X     [1,1,0,1],[127X[104X
    [4X[25X>[125X [27X     [1,1,1,0]] );  # inefficient way to compute NrDerangements([1..4])[127X[104X
    [4X[28X9[128X[104X
    [4X[25Xgap>[125X [27X# 24 permutations fit the projective plane of order 2:[127X[104X
    [4X[25Xgap>[125X [27XPermanent( [[1,1,0,1,0,0,0],[127X[104X
    [4X[25X>[125X [27X     [0,1,1,0,1,0,0],[127X[104X
    [4X[25X>[125X [27X     [0,0,1,1,0,1,0],[127X[104X
    [4X[25X>[125X [27X     [0,0,0,1,1,0,1],[127X[104X
    [4X[25X>[125X [27X     [1,0,0,0,1,1,0],[127X[104X
    [4X[25X>[125X [27X     [0,1,0,0,0,1,1],[127X[104X
    [4X[25X>[125X [27X     [1,0,1,0,0,0,1]] );[127X[104X
    [4X[28X24[128X[104X
  [4X[32X[104X
  
