Question
In how many ways can the letters of the word “MATHEMATICS” be arranged so that:
(a) All vowels come together? (b) No two vowels are adjacent?
Solution — Step by Step
MATHEMATICS has 11 letters: M, A, T, H, E, M, A, T, I, C, S.
Frequencies: M appears 2 times, A appears 2 times, T appears 2 times. Vowels: A, E, A, I (4 letters, with A repeated).
Consonants: M, T, H, M, T, C, S (7 letters, with M and T each repeated twice).
Bundle all 4 vowels into a single super-letter. Now we arrange 7 consonants + 1 vowel-block = 8 units, with M and T each repeated twice.
Arrangements of these 8 units:
Inside the vowel block, the 4 vowels (A, A, E, I) can be arranged in:
Total: .
First, arrange the 7 consonants. They have 7 positions plus 8 “gaps” (including the two ends) where vowels can be slotted.
Consonant arrangements:
Choose 4 of the 8 gaps for the vowels (so no two vowels share a gap, hence no two are adjacent):
Arrange the 4 vowels in those 4 chosen gaps:
Total: .
Final: (a) 120960, (b) 1058400.
Why This Works
The “block trick” handles “together” conditions cleanly — by gluing the items together first, then permuting within the glued block. The “gaps method” handles “no two adjacent” conditions — place the constrained items first, then slot the remaining items into the gaps.
Repeated letters always require dividing by the factorial of each repeat count. Forgetting this is the most common P&C error in CBSE board exams.
Alternative Method
For (b), use inclusion-exclusion: subtract from total arrangements the cases where at least two vowels are adjacent. Vastly more cumbersome — the gaps method is the standard.
Common Mistake
Students forget to permute the vowels within their block in (a). They report instead of . Whenever you bundle items into a block, remember that the bundle’s internal arrangement also contributes to the count.