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Find the number of distinct arrangements of all the letters in the word 'MISSISSIPPI'.
A$11!$
B$11!/(4!\,4!\,2!)$
C$\dfrac{11!}{2!}$
D$11^4$
Answer & Solution
Correct answer: B. $11!/(4!\,4!\,2!)$
1. MISSISSIPPI has 11 letters: M(1), I(4), S(4), P(2).
2. Total letters: $11$. With ALL DISTINCT, the count would be $11!$.
3. But there are REPEATED letters. For each repeated letter group of size $k$, we OVERCOUNT by $k!$.
4. Adjust: divide by $4!$ (for the four I's), $4!$ (four S's), $2!$ (two P's), $1!$ (one M — trivial).
5. Result: $\dfrac{11!}{4!\,4!\,2!\,1!}$. Numerically: $39{,}916{,}800 / (24 \cdot 24 \cdot 2 \cdot 1) = 39{,}916{,}800/1152 = 34{,}650$.
6. Option A treats all letters as distinct (over-counts). Option C only adjusts for P's. Option D is unrelated.
_Source: NCERT Class 11 Mathematics, Ch 6, §6.3.2 (Permutations with repetition), p. 7._
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