Suppose there is a word  abcde which has all the alphabets distinctly different. Then no. of ways in which the alphabets can be arranged is 5! .if suppose one alphabet is repeated twice, say in abace, then no. of ways alphabets can be arranged is (5! / 2!). if suppose one alphabet is repeated thrice, say in abaca, then no. of ways alphabets can be arranged is (5! / 3!).  If there are m different alphabets and only one alphabet is repeated n times, then no. of arrangements is (m!/n!). Suppose there are m no. of alphabets out of which one alphabet is repeated n1 times, another alphabet is repeated n2 times, another alphabet is repeated n3 times, so on so forth (n1+n2+n3+......=m), then total no. of arrangements that can be made is m!/(n1!n2!n3!......). Here n1,n2,n3,........ are called Partition Functions. Exa-1:--ABCBACBACBCDED . Here m=14, n1(A)=3,n2(B)=4,n3(C)=4,n4(D)=2,n5(E)=1 and n1+n2+n3+n4+n5=m=14. No. of ways the alphabets in the word  can be arranged is m!/(n1!n2!n3!n4) = 14!/(3!4!4!2!1!). Here there are 4 partition functions. Exa-2: No. of ways the word PROBABILITY can be arrangement is 11!/(1!1!1!2!1!2!1!1!1!)=(11!/2!2!) where m=11,   n1(P) =1,n2(R)=1,n3(O)=1,n4(B)=2,n5(A)=1,n6(I)=2,n7(L)=1,n8(T)=1,n9(Y)=1.  here there are 9 partition functions n1,n2 ,....n9 and their sum is 11 i.e. equal to m.

If all or any of the n1,n2,n3, .....n6 alphabets will be always together, then m is modified to m1=m-[(n1t-1)+(n2t-1)+(n3t-1)+(n4t-1)+(n5t-1)+(n6t-1)+(n7t-1)+(n8t-1)+(n9t-1)+(n10t-1)] =m- Σ(nit-1) where i=1 to 10 & nit=1 if ni=0 and nit=ni if ni !=0 and no. of arrangements become m1!/(n1r!n2r!n3r!n4r!n5r!n6r!n7r!n8r!n9r!n10r) where nir=ni when nit=1 otherwise nir=1 & i=1 to 10 here.