IPMAT Indore 2022 · Quantitative Ability · Question 22

Step 1 — inventory the letters. MANAGEMENT has 10 letters:

Consonants (6): M, N, G, M, N, T — with M twice and N twice

Vowels (4): A, A, E, E — with A twice and E twice

Step 2 — arrange the consonants first.

\frac{6!}{2!\,2!}=\frac{720}{4}=180

Step 3 — choose the gaps for the vowels. Six consonants in a row create 7 gaps (including the two ends):

\_C \_C \_C \_C \_C \_C \_

Placing at most one vowel per gap guarantees no two vowels are adjacent. Choose 4 of the 7 gaps and arrange the vowels (which have repeats):

\left( \genfrac{}{}{0pt}{}{7}{4}\right)\times \frac{4!}{2!\,2!}=35\times 6=210

Step 4 — multiply.

180\times 210=37800

Answer: Option 3 — 37800

Shortcut. Gap method, and divide by the repeats on both halves.

\underset{\text{consonants}}{\underset{⏟}{\frac{6!}{2!2!}}}\times \underset{\text{vowels into gaps}}{\underset{⏟}{\left( \genfrac{}{}{0pt}{}{7}{4}\right)\frac{4!}{2!2!}}}=180\times 210=37800

Two slips to avoid, each of which matches a wrong option:

Forgetting the vowel repeats gives 180\times 35\times 24=151200;

Using n gaps instead of n+1 (6 rather than 7) gives 180\times 15\times 6=16200.

Six letters in a row give seven gaps — count the ends. That is the single most common error in “no two together” questions.