ARRANGEMENT OF ALPHABETS

 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. No. of Alphabets (m) *Partition function:n1 Partition function:n2 Partition function:n3 Partition function:n4 Partition function:n5 Partition function:n6 Partition function:n7 Partition function:n8 Partition function:n9 Partition function:n10 Whether alphabets with partition function n1 will *always be together ?(n1t) no yes Whether alphabets with partition function n2 will always be together ?(n2t) no yes Whether alphabets with partition function n3 will always be together ?(n3t) no yes Whether alphabets with partition function n4 will always be together ?(n4t) no yes Whether alphabets with partition function n5 will always be together ?(n5t) no yes Whether alphabets with partition function n6 will always be together ?(n6t) no yes Whether alphabets with partition function n7 will always be together ?(n7t) no yes Whether alphabets with partition function n8 will always be together ?(n8t) no yes Whether alphabets with partition function n9 will always be together ?(n9t) no yes Whether alphabets with partition function n10 will always be together ?(n10t) no yes * means they may be or may not be together   ** sum of partition functions should be equal to m. m1 no. of arrangements if all alphabets were distinctly different: m! no. of arrangements: Example: If the letters of the word  PROBABILITY are written down at random in a row, the probability that two B's are together is Ans- Here 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 . Total no. of arrangements=9979200, If B's are together, n4t=yes and total no. of arrangements=1814400 . Dividing, we get 2/11. Example: The letters of the word COCHIN are permuted and all the permutations are arranged in alphabetical order as in English dictionary. The no. of words that appear before the word COCHIN is Ans: COCHIN can be arranged so that words with CC.....,CH....,CI....,CN.... appear before COCHIN. Leaving first 2 alphabets, other 4 can be arranged in 4!=24 ways. Since there are 4 of them , total 4*24=96 words appear before CO......