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
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)
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.
. Total no. of arrangements=9979200, If B's are together, n4t=yes
and total no. of arrangements=1814400 . Dividing, we get 2/11.