IPMAT Indore 2024 · Quantitative Ability · Question 21

Reading the condition. “At least one boy between any two girls” means no two girls are adjacent.

Step 1 — total arrangements. Eight people around a circular table:

\left( 8-1 \right)!=7!=5040

Step 2 — favourable arrangements. Seat the boys first, then slot the girls into the gaps.

Arrange the 5 boys in a circle: \left( 5-1 \right)!=4!=24 ways.

Five boys create 5 gaps between them. Choose 3 gaps for the 3 girls and order the girls:

{}^{5}{P}_{3}=5\times 4\times 3=60

Placing at most one girl per gap guarantees no two girls are adjacent.

\text{Favourable}=24\times 60=1440

Step 3 — divide.

P=\frac{1440}{5040}=\frac{2}{7}

Answer: Option 1 — \frac{2}{7}

Shortcut. Gap method, then cancel early.

P=\frac{4!\times{}^{5}{P}_{3}}{7!}=\frac{4!\cdot 5\cdot 4\cdot 3}{7!}

Write 7!=7\cdot 6\cdot 5! and note 4!\cdot 5=5!, so

P=\frac{5!\cdot 12}{7\cdot 6\cdot 5!}=\frac{12}{42}=\frac{2}{7}

The circular-gap rule: n people in a circle make exactly n gaps (not n+1 as in a row). Using n+1 is the single commonest error here and leads to 1/3.