Question

In how many of the distinct permutations of the letters in $MISSISSIPPI$ do the four $I^{\prime }s$ not come together?

Answer 1

Total number of letters in $MISSISSIPPI=11$
There are $4I^{\prime }s,4S^{\prime }s,2P^{\prime }s$ (repetition)
Total numebr of permutations $=\frac{n!}{P_1!P_1!P_1!}$
$=\frac{11!}{4!4!2!}=\frac{11\times 10\times 9\times 8\times 7\times 6\times 5\times 4!}{(4\times 3\times 2\times 1)\times (4!)\times (2\times 1)}=34650$

Taking all $I^{\prime }s$ together.
$I^{\prime }s$ in $MISSISSIPPI=4$
$\because$ All $I^{\prime }s$ occur together, so assume $IIII$ as single letter.
$\therefore$ Letters become $M,S,S,S,S,P,P,IIII\to 8$ letters
And $S$ is repeating $4$ times and $P$ is repeating $2$ times.


Number of Permutation of $8$ letters.
$\frac{{}^8{P}_8}{4!\times 2!}=\frac{8!}{4!\times 2!}$
$=\frac{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{4\times 3\times 2\times 1\times 2\times 1}=840$

The number of distinct permutations of the letters in $MISSISSIPPI$ when the four $I^{\prime }s$ not come together $=$ Total arrangments $-$ arrangements when $4I^{\prime }s$ come together
$=34650-840$
$=34650-840=33810$.