Given,
The word ASSASSINATION.
It has 13 letters of which repeated letters are A, S, I, and N repeating thrice, 4 times, twice and twice respectively.
We have to find number of words that can formed in such a way that all S’s must come together.
For example -
ASSSSANNIITO, ININAASSSSATO are few words among them.
Notice that all S letters comes in bunch (In bold letters).
A specific method is usually used for solving such type of problems.
According to that we assume the group of letters that are remains together (here S, S, S, S) are assumed to be a single letter and all other letters are as usual counted as single letter.
Now find number of ways as usual;
Number of ways of arranging r letters from a group of n letters is equals to nPr .
And the final answer is then multiplied by number of ways we can arrange the letters in that group which has to be sticked together in it (Here S, S, S, S).
Since we know,
Permutation of n objects taking r at a time is nPr,and permutation of n objects taking all at a time is n!
And,
We also know,
Permutation of n objects taking all at a time having p objects of same type, q objects of another type, r objects of another type is \(\frac{n!}{p!\times q!\times r!}\).
i.e.,
The number of repeated objects of same type are in denominator multiplication with factorial.
Now,
Letters of word ASSASSINATION : 13
Now from our method,
Letters are SSSS, A, A, A, I, I, N, N, T, O. (10 letters)
Total number of arrangements of 10 letters = \(\frac{10!}{3!\times 2!\times 2!}\)
And total number of arrangements of grouped letters (Here S, S, S, S) is \(\frac{4!}{4!}\),equals 1.
So, our final answer for arranging the letters such that all vowels sticks together equals multiplication of \(\frac{10!}{3!\times 2!\times 2!}\) and \(\frac{4!}{4!}\).
Total number of arrangements = \(\frac{10!}{3!\times 2!\times 2!}\)\(\times\)\(\frac{4!}{4!}\)
= 151200.
Hence,
Total number of words formed during arrangement of letters of word ASSASSINATION such that all S’s remains together is equals to 151200.
