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In how many of the distinct permutations of the letters in MISSISSIPPI do the four I's not come together?
Solution:
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Verified Answer
In given word there are $4 \mathrm{I}, 4 \mathrm{~S}, 2 \mathrm{P}$ and $1 \mathrm{M}$.
Total number of permutations $=\frac{11 !}{4 ! 4 ! 2 !}$
If take 4 I's as one letter then total letters become $=11-4+1=8$
If $\mathrm{P}$ is the permutations when 4 I's are not together, then
$\begin{aligned}
&P=\frac{11 !}{4 ! 4 ! 2 !}-\frac{8 !}{4 ! 2 !} \\
&=\frac{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{4 \times 3 \times 2 \times 1 \times 2 \times 1 \times 4 !}-\frac{8 \times 7 \times 6 \times 5 \times 4 !}{2 \times 1 \times 4 !} \\
&=34650-840=33810
\end{aligned}$
Total number of permutations $=\frac{11 !}{4 ! 4 ! 2 !}$
If take 4 I's as one letter then total letters become $=11-4+1=8$
If $\mathrm{P}$ is the permutations when 4 I's are not together, then
$\begin{aligned}
&P=\frac{11 !}{4 ! 4 ! 2 !}-\frac{8 !}{4 ! 2 !} \\
&=\frac{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{4 \times 3 \times 2 \times 1 \times 2 \times 1 \times 4 !}-\frac{8 \times 7 \times 6 \times 5 \times 4 !}{2 \times 1 \times 4 !} \\
&=34650-840=33810
\end{aligned}$
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