How many words can be formed from MISSISSIPPI

Question

How many distinct arrangements (words) can be formed using all the letters of the word MISSISSIPPI?

Solution — Step by Step

MISSISSIPPI has 11 letters. Let’s count each letter:

M: 1
I: 4
S: 4
P: 2

Total: 1 + 4 + 4 + 2 = 11 letters ✓

When we arrange $n$ objects where some are identical, the number of distinct arrangements is:

\text{Number of arrangements} = \frac{n!}{n_1! \times n_2! \times n_3! \times \ldots}

where $n_1, n_2, n_3, \ldots$ are the frequencies of each repeated element.

Why? Because if we had $n!$ total arrangements (treating all as distinct), we’ve over-counted. The $n_1!$ identical objects can be rearranged among themselves in $n_1!$ ways without changing the overall arrangement — so we divide by $n_1!$ to eliminate these duplicates.

\text{Number of words} = \frac{11!}{1! \times 4! \times 4! \times 2!}

Now compute:

11! = 39916800

4! = 24, \quad 4! = 24, \quad 2! = 2, \quad 1! = 1

Denominator: $1 \times 24 \times 24 \times 2 = 1152$

\text{Number of words} = \frac{39916800}{1152} = 34650

\frac{39916800}{1152}

$39916800 \div 1152$ : First, $1152 \times 30000 = 34560000$ . Remaining: $39916800 - 34560000 = 5356800$ .

$1152 \times 4000 = 4608000$ . Remaining: $5356800 - 4608000 = 748800$ .

$1152 \times 600 = 691200$ . Remaining: $748800 - 691200 = 57600$ .

$1152 \times 50 = 57600$ . ✓

Total: $30000 + 4000 + 600 + 50 = 34650$ .

The number of distinct arrangements of MISSISSIPPI = 34,650.

Why This Works

If all 11 letters were distinct, we’d have $11!$ arrangements. But since I appears 4 times, the 4 I’s can be swapped among their positions without creating a new word — that gives $4! = 24$ ways we over-counted for I’s. Similarly for S’s (over-count by 24) and P’s (over-count by 2).

Dividing by the factorials of the repeated letter counts removes all these duplicates, leaving only genuinely distinct arrangements.

Alternative Method — Build up from smaller factorials

To avoid computing full factorials, simplify the fraction first:

\frac{11!}{4! \cdot 4! \cdot 2!} = \frac{11!}{4! \cdot 4! \cdot 2}

Work step by step using combination-style simplification:

= \binom{11}{4} \times \binom{7}{4} \times \binom{3}{2} \times 1!

= \frac{11!}{4! \cdot 7!} \times \frac{7!}{4! \cdot 3!} \times \frac{3!}{2! \cdot 1!}

= 330 \times 35 \times 3 = 34650

This approach uses smaller numbers and is less prone to arithmetic errors.

CBSE Class 11 and JEE Main both test this “arrangement with repetition” formula. The standard examples are MATHEMATICS (11 letters: M×2, A×2, T×2 → $\frac{11!}{2!2!2!}$ ), MISSISSIPPI, and BANANA. For BANANA: B×1, A×3, N×2 → $\frac{6!}{3! \cdot 2!} = 60$ .

Common Mistake

Students forget to account for ALL repeated letters. For MISSISSIPPI, you must divide by $4!$ (for I), $4!$ (for S), AND $2!$ (for P). Forgetting P gives $\frac{11!}{4! \times 4!} = 69300$ — exactly double the correct answer. Check: count unique letters first, then check which have frequency > 1.

Question

Solution — Step by Step

Why This Works

Alternative Method — Build up from smaller factorials

Common Mistake

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