6-Letter Words From MATHS Where Every Letter Appears Twice

JEE Main 2025 — 29 January, Morning Shift. Previous Year Question.

Problem

The number of 6-letter words, with or without meaning, that can be formed using the letters of the word MATHS such that any letter that appears in the word must appear at least twice, is _.

Key insight. MATHS has only 5 distinct letters, so a 6-letter word must repeat at least one of them — but the condition is stricter: every letter used has to appear twice or more. That turns the problem into counting the ways to split 6 positions into groups of size 2 or more, one group per letter chosen.

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Approach

Since MATHS contributes 5 distinct letters and the word has 6 positions, at least one letter must repeat. The condition “any letter used must appear at least twice” rules out ever using a letter only once, which means the only thing to figure out is how the number 6 can be split into parts of size 2 or more — each part being how many times one particular letter is used. Once the possible splits (partitions) are listed, each one becomes a separate counting problem: choose which letters play which role, then count the arrangements of that multiset.

Solution

Step 1 — List the ways 6 can split into parts of size ≥ 2

Since no letter can appear just once, every valid word corresponds to writing 6 as a sum of parts each at least 2, where each part is the count of one distinct letter used:

$$6 = 6, \quad 6 = 4+2, \quad 6 = 3+3, \quad 6 = 2+2+2$$

These are the only four ways — a split like $2+2+1+1$ is invalid because parts of size 1 aren’t allowed.

Step 2 — One letter, six times

Choose 1 letter out of the 5 available and repeat it six times. There’s nothing left to arrange since every position is identical:

$$\binom{5}{1} \times 1 = 5$$

Step 3 — One letter four times, another twice

Here the two chosen letters play different roles, and swapping who gets 4 and who gets 2 produces genuinely different words — a word with four A’s and two M’s is not the same as one with four M’s and two A’s. So this case is counted twice: once for each assignment of roles.

For a fixed assignment, choose the 2 letters out of 5, then arrange the six positions with one letter repeated 4 times and the other repeated twice:

$$\binom{5}{2} \times \frac{6!}{4!\,2!} = 10 \times 15 = 150$$

Since the roles can be assigned in 2 ways (either letter could be the one used four times):

$$2 \times 150 = 300$$

Step 4 — Two letters, three times each

Here the two letters play symmetric roles, so there’s no double-counting to worry about:

$$\binom{5}{2} \times \frac{6!}{3!\,3!} = 10 \times 20 = 200$$

Step 5 — Three letters, two times each

Choose 3 of the 5 letters, then arrange six positions with each of the three letters repeated twice:

$$\binom{5}{3} \times \frac{6!}{2!\,2!\,2!} = 10 \times 90 = 900$$

Step 6 — Add up every case

$$5 + 300 + 200 + 900 = 1405$$

Answer

Common mistakes

• Forgetting that the “4 and 2” case is really two cases. It’s tempting to compute $\binom{5}{2} \times \frac{6!}{4!\,2!}$ once and stop, but a word with four A’s and two M’s is a different word from one with four M’s and two A’s. Missing this halves that part of the answer.
• Including invalid splits. A split like $2+2+1+1$ looks like it uses six positions too, but it puts one letter in just once, which the question explicitly disallows.
• Treating repeated letters as if position order doesn’t matter. The multinomial arrangement formula $\frac{6!}{a!\,b!\,\cdots}$ already accounts for indistinguishable repeats, so there’s no need for a separate arrangement step.

Practise next

• Count the 5-letter words that can be formed from the four letters of LOVE, where every letter used must appear at least twice. The same partition technique works, on a smaller case.