Question

Consider the word $MISSISSIPPI$

If the number of arrangment of the letters of the word $W$ if all the $S^{\prime }s$ and $P^{\prime }$ are seperated is $\left(K\right)$ $\left(\frac{10!}{4!.4!}\right)$ then $K$ equals

• A. $\frac{6}{5}$
• B. $1$
• C. $\frac{4}{3}$
• D. $\frac{3}{2}$

Answer 1

The correct option is B $1$

$W=MISSISSIPPIM-1,I-4,S-4,P-2$

$Leaving{}^{\prime }{S}^{\prime }.$

$Numberofarragements:\frac{7!}{4!2!}=105$

$I{f}^{\prime }{P}^{\prime }istogetherNumberofarrangements:\frac{6!}{4!}=30$

$Numberofwaysifinserting{}^{\prime }{S}^{\prime }:8_{C_4}$

$NumberofwaysinwhichPandSbothareseparated=30\cdot 7_{C_3}+75\cdot 8_{C_4}=6300$

According to question,

$K\left(\frac{10!}{4!4!}\right)=6300K=6300\times \frac{4!\times 4\times 3\times 2\times 1}{10\times 9\times 8\times 7\times 6\times 5\times 4!}$

$\therefore K=1$

$Hence,optionBiscorrectanswer.$