Permutations and Combinations: Speed-Solving Techniques (10)

easy 2 min read

Question

In how many ways can the letters of the word “MISSISSIPPI” be arranged? Of these arrangements, how many have all four S’s together? JEE Main 2024 pattern.

Solution — Step by Step

Total letters: 11. Frequencies: M = 1, I = 4, S = 4, P = 2.

Total arrangements with repetition adjustments:

N=11!1!4!4!2!N = \frac{11!}{1! \cdot 4! \cdot 4! \cdot 2!}

Compute: 11!=39,916,80011! = 39{,}916{,}800. 4!4!2!=24242=11524! \cdot 4! \cdot 2! = 24 \cdot 24 \cdot 2 = 1152.

N=39,916,8001152=34,650N = \frac{39{,}916{,}800}{1152} = 34{,}650

Glue the four S’s into one super-letter. Now we have 8 entities to arrange: M, I, I, I, I, P, P, and the SSSS block.

Arrangements of 8 entities with repetitions:

NS together=8!1!4!2!1!=4032048=840N_{\text{S together}} = \frac{8!}{1! \cdot 4! \cdot 2! \cdot 1!} = \frac{40320}{48} = 840

Total arrangements: 34,65034{,}650. Arrangements with all S’s together: 840840.

Why This Works

The formula n!n1!n2!nk!\dfrac{n!}{n_1! n_2! \cdots n_k!} counts permutations of nn objects when there are groups of identical objects. We divide by the factorials of group sizes because permuting identical objects within a group doesn’t produce a new arrangement.

The “all together” trick is universal: glue them into a single entity, count arrangements, and (if those entities are distinguishable internally) multiply by the internal arrangements. Here all four S’s are identical, so internal arrangements contribute a factor of 1.

Alternative Method

For “S’s together”, you could use the gap method: arrange the other 7 letters (M, I, I, I, I, P, P) in 7!4!2!=105\dfrac{7!}{4! \cdot 2!} = 105 ways, then place the SSSS block in any of the 8 gaps. Total: 105×8=840105 \times 8 = 840. Same answer.

Students often forget to divide by the factorial of the count of each repeated letter. With four I’s, four S’s, and two P’s, you must divide by 4!4!2!4! \cdot 4! \cdot 2!, not just 4!4!.

For “always together” problems: glue the special letters into a block, then count arrangements of the new (smaller) set. For “never together” problems: use total arrangements minus “together” arrangements, or use the gap method.

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