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 \underline\hspace2cm .

Permutations and Combinations PYQ — JEE Main 2025 · 29 Jan · Shift 1 · Q50 with Solution

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Correct answer: 1405

The word MATHS has 5 distinct letters: $\{M,A,T,H,S\}$
We need 6-letter words formed from these letters such that:
- repetition is allowed,
- but if a letter appears, it must appear at least twice.
So, in a 6-letter word, the multiplicities of used letters must be a partition of 6 into parts each at least 2.

Possible distributions are: $6,\,4+2,\,3+3,\,2+2+2$

Now count each case.

Case 1: One letter repeated 6 times

Choose the letter: $\binom{5}{1}=5$ Only one arrangement for each choice.

So count = $5$

Case 2: Distribution $4+2$

Choose 2 letters out of 5: $\binom{5}{2}=10$ From these 2 letters, choose which one appears 4 times: $2$ ways Arrange the multiset $(4,2)$ in 6 positions: $\frac{6!}{4!2!}=15$

Total for this case:

=10\cdot 2\cdot 15=300$$ --- ### Case 3: Distribution $3+3$ Choose 2 letters out of 5: $$\binom{5}{2}=10$$ Arrange 3 of one and 3 of the other: $$\frac{6!}{3!3!}=20$$ Total for this case: $$10\cdot 20=200$$ --- ### Case 4: Distribution $2+2+2$ Choose 3 letters out of 5: $$\binom{5}{3}=10$$ Arrange the multiset: $$\frac{6!}{2!2!2!}=90$$ Total for this case: $$10\cdot 90=900$$ --- ### Total Add all cases: $$5+300+200+900=1405$$ Therefore, the required number of words is $$\boxed{1405}$$ --- ### Comparison with stored answer Stored correct answer = **1405**. Our derived answer is **1405**, so it agrees.

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View written solutionFree

Correct answer: 1405

The word MATHS has 5 distinct letters: $\{M,A,T,H,S\}$
We need 6-letter words formed from these letters such that:
- repetition is allowed,
- but if a letter appears, it must appear at least twice.
So, in a 6-letter word, the multiplicities of used letters must be a partition of 6 into parts each at least 2.

Possible distributions are: $6,\,4+2,\,3+3,\,2+2+2$

Now count each case.

Case 1: One letter repeated 6 times

Choose the letter: $\binom{5}{1}=5$ Only one arrangement for each choice.

So count = $5$

Case 2: Distribution $4+2$

Choose 2 letters out of 5: $\binom{5}{2}=10$ From these 2 letters, choose which one appears 4 times: $2$ ways Arrange the multiset $(4,2)$ in 6 positions: $\frac{6!}{4!2!}=15$

Total for this case:

=10\cdot 2\cdot 15=300$$ --- ### Case 3: Distribution $3+3$ Choose 2 letters out of 5: $$\binom{5}{2}=10$$ Arrange 3 of one and 3 of the other: $$\frac{6!}{3!3!}=20$$ Total for this case: $$10\cdot 20=200$$ --- ### Case 4: Distribution $2+2+2$ Choose 3 letters out of 5: $$\binom{5}{3}=10$$ Arrange the multiset: $$\frac{6!}{2!2!2!}=90$$ Total for this case: $$10\cdot 90=900$$ --- ### Total Add all cases: $$5+300+200+900=1405$$ Therefore, the required number of words is $$\boxed{1405}$$ --- ### Comparison with stored answer Stored correct answer = **1405**. Our derived answer is **1405**, so it agrees.

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