Question

Total number of words formed by the letters of the word “MISSISSIPPI” in which any two $S$are separated is equal to:-
A. $7350$
B. $6300$
C. $12600$
D. $5000$

Answer 1

Hint: As we know that the above question is of permutations. We can see that in the word MISSISSIPPI there are four I’s, four S’s, two P’s and one M. We have to find the number of ways the word can be arranged. Using the concept of permutations we can identify the formula and get the number of ways. The basic formula that can be applied in permutations is $\dfrac{{n!}}{{{p_1}!{p_2}!{p_3}!...}}$.

Complete step by step answer:
We can see that in the word MISSISSIPPI there are total $11$ alphabets i.e. $n = 11$ which they can be arranged in total $11$ ways. It means the number of possible ways of $11$ alphabets is $11!$.
But as we can see that there are $4I,4S,2P$, they all can be arranged in $4!,4!,2!$, since these are repeated letters. Also the question says that no two $S$should be together, so we can say it the word arrangement is: $M\_I\_I\_I\_I\_P\_P\_$. It can be arranged in $7$ways i.e.$7!$.
Therefore the possible number of words is given by $\dfrac{{^8{C_4} \times 7!}}{{4!2!}}$.
 By substituting the values in the formula the number of arrangements possible with the given alphabets are $\dfrac{{8! \times 7!}}{{4! \times 4! \times 4! \times 2!}}$, simplifying this we get $\dfrac{{8 \times 7 \times 6 \times 5 \times 4! \times 7 \times 6 \times 5 \times 4!}}{{4! \times 4! \times 4 \times 3 \times 2 \times 2 \times 1}} = \dfrac{{8 \times 7 \times 6 \times 5 \times 7 \times 6 \times 5}}{{4 \times 3 \times 2 \times 1 \times 2 \times 1}}$It gives the number of ways which is $7350$.

Hence the total number of possible word formed by MISSISSIPPI keeping no S together are $(a)7350$.

Note: We should know that permutations means arrangement which is simply the grouping of objects according to the requirement. We should be careful while calculating the number of ways in the word since there are few letters that are repeated and they need to be calculated together. Whenever we face such types of problems the value point to remember is that we need to have a good knowledge of permutations and its formulas.