How Many Different Words Can Be Formed by Jumbling the Letters of the Word MISSISSIPPI in Which No Two S’s Are Adjacent

AIEEE 2008 — Previous Year Question.

Problem

How many different words can be formed by jumbling the letters of the word MISSISSIPPI, in which no two S’s are adjacent?

Key insight. Rather than trying to place all 11 letters at once while somehow keeping the S’s apart, arrange every other letter first — that creates a fixed number of gaps between them, and dropping the S’s into distinct gaps is what automatically guarantees none of them end up adjacent.

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Approach

MISSISSIPPI has 11 letters: M (1), I (4), S (4), P (2). Set the four S’s aside for a moment and arrange the remaining 7 letters (M, I×4, P×2) among themselves. This creates 8 gaps — one before the first letter, one after the last, and one between each pair of consecutive letters — into which the 4 identical S’s can be placed one per gap, which is exactly what keeps every pair of S’s separated by at least one other letter.

Solution

Step 1 — Count the letters excluding S

Removing the 4 S’s leaves 7 letters: M (1), I (4), P (2).

Step 2 — Arrange these 7 letters

Since I repeats 4 times and P repeats 2 times, the number of distinct arrangements is:

$$\frac{7!}{4!\,2!}$$

This can be rewritten to match the answer choices’ form: $7!=7\times6!$, and $\dfrac{6!}{4!\,2!}={}^6C_4$ (choosing 4 of the 6 remaining positions for the I’s, say), so:

$$\frac{7!}{4!\,2!} = 7\times{}^6C_4$$

Step 3 — Count the gaps created

Arranging 7 letters in a row creates 8 possible gaps for inserting other letters: one before the first letter, one between each of the 6 consecutive pairs, and one after the last letter — that’s $6+2=8$ gaps in total.

Step 4 — Place the 4 S’s into distinct gaps

To ensure no two S’s are adjacent, each S must go into a different gap (placing two S’s in the same gap would put them next to each other). Choosing which 4 of the 8 gaps receive an S:

$${}^8C_4$$

Since the S’s are identical, there’s no further arrangement needed once the gaps are chosen — each selected gap simply gets one S.

Step 5 — Combine both counts

$$\text{Total arrangements} = \left(7\times{}^6C_4\right)\times{}^8C_4 = 7\times{}^6C_4\times{}^8C_4$$

Answer

Common mistakes

• Trying to arrange all 11 letters directly, then subtracting the “bad” arrangements with adjacent S’s. This inclusion-exclusion approach is far messier than the gap method here, since there are multiple ways for S’s to clump together (2 together, 3 together, all 4 together) that would each need separate handling.
• Counting 7 gaps instead of 8. It’s easy to only count the gaps between the 7 letters (6 of them) and forget the two end positions (before the first letter and after the last) — both ends are valid places for an S too.
• Multiplying by an extra arrangement factor for the S’s. Since all 4 S’s are identical, once the 4 gaps are chosen there’s exactly one way to fill them — there’s no ${}^{4}P_4$ or $4!$ factor to include, unlike arranging distinct objects into chosen slots.

Practise next

• How many different words can be formed by jumbling the letters of ALABAMA (A×4, L, B, M) such that no two A’s are adjacent?