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Question: How many different words can be formed by jumbling the letter in the word MISSISSIPPI in which no tw...

How many different words can be formed by jumbling the letter in the word MISSISSIPPI in which no two S are adjacent?
A. 7×6C4×8C47{ \times ^6}{C_4}{ \times ^8}{C_4}
B. 8×6C4×7C48{ \times ^6}{C_4}{ \times ^7}{C_4}
C. 6×7×8C46 \times 7{ \times ^8}{C_4}
D. 6×8×7C46 \times 8{ \times ^7}{C_4}

Explanation

Solution

We will first of all calculate all the letters individually. Now, we will first remove all the S’s from the letters provided to us and write the other letters leaving one space in between them where we are going to arrange S’s.

Complete step-by-step solution:
We have the word MISSISSIPPI, where we have 1 M, 4 I’s, 4 S’s and 2 P’s.
Now, if we remove the S’s for once, then we are left with 1 M, 4 I’s and 2 P’s.
Writing the word given without S’s, we will get:-
__ M __ I __ I __ I __ P __ P __ I __
Now, the blanks in between are the places where we can arrange the S’s and there cannot be any S together by that.
Now, we have 8 blanks and 4 S’s. So, we need to choose 4 S from 8 S’s.
There are 8C4^8{C_4} ways to do so because to choose r objects from n objects, we have nCr^n{C_r} ways. …….(1)
Now, we just need to permute the rest of words which are 7 letters therefore can be permuted in 7! Ways but since there are 4 I’s and 2 P’s.
Therefore, we can permute the letters in 7!2!×4!\dfrac{{7!}}{{2! \times 4!}} ways.
We can write it as: 7×6!2!×4!7 \times \dfrac{{6!}}{{2! \times 4!}}.
Now, since we know that nCr=n!r!(nr)!^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}.
Taking n = 6 and r = 4 in it, we will get:-
7×6C4\Rightarrow 7{ \times ^6}{C_4} ………………(2)
Combining (1) and (2), we will get:-
Total number of ways required = 7×6C4×8C47{ \times ^6}{C_4}{ \times ^8}{C_4}

\therefore The correct option is (A).

Note: The students must note that, to permute the letters which have some identical letters in between we need to divide it as we did in this question, because if we do not divide it, we will get some same arrangement multiple times which will hamper the answer we get.
Permutation and Combination is such a relief and important topic because, if we would have tried to write down all the words to calculate them, it would have taken an eternity to do so. We calculated the number of ways so easily now.