Question
In how many ways can the letters of the word MATHEMATICS be arranged such that all vowels come together? Then find the number of arrangements where no two vowels are adjacent. This pair of “all together vs none together” questions is a JEE Main staple.
Solution — Step by Step
MATHEMATICS has letters:
- M appears times
- A appears times
- T appears times
- H, E, I, C, S each time
Vowels: A, A, E, I — that’s vowels.
Consonants: M, M, T, T, H, C, S — that’s consonants.
Treat all vowels as a single super-letter. Then we have consonants vowel-bundle = items.
Arrangements of these items (with M repeated twice and T repeated twice):
Within the vowel-bundle, vowels can be arranged with A repeated:
Total: .
First, arrange the consonants (M, M, T, T, H, C, S):
Now place vowels in the gaps. With consonants in a row, there are gaps (one before, six between, one after):
Choose gaps out of for the vowels: ways. Then arrange the vowels (A, A, E, I) in those chosen positions:
Total: .
Why This Works
Together-trick (bundling): When a problem says ” items must be together”, treat them as a single unit, count arrangements of the reduced set, then multiply by internal arrangements within the bundle. This is permutation algebra at its cleanest.
Apart-trick (gap method): When items must not be adjacent, arrange the others first to create gaps, then choose gaps to place the restricted items. Always count gaps as (number of items) + 1.
The interplay between these two tricks is what JEE tests — many problems combine them (“all vowels together but consonants in a specific order”, etc.).
Alternative Method
For “no two vowels adjacent”, you can also use complementary counting: total arrangements minus arrangements with at least two vowels adjacent. Inclusion-exclusion gets messy here because of multiple vowel pairs, so the gap method is much cleaner.
Always check the "" in the gap count. With items in a row, there are gaps (including the two ends). Forgetting the end-gaps gives only , which silently halves your answer in some problems.
Common Mistake
Three classic traps in this template:
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Forgetting to divide by repeats inside the vowel bundle. Vowels A, A, E, I have only distinct internal arrangements, not . Always check for repeated letters in every permutation count.
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Using alone, forgetting to arrange the vowels. Choosing the gaps just places the vowels — they still need to be arranged among themselves. Multiply by the internal arrangement count.
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Counting both end-gaps correctly. Some students think consonants give only internal gaps and miss the two ends. The right count is always gaps for consonants.
Final answer: Vowels together: . No two vowels adjacent: .