JEE Main 2025 Jan 22 Shift 2, Mathematics Q2: Permutation

JEE Main2025Jan 22, Shift 2Mathematics

In a group of 3 girls and 4 boys, there are two boys $B_{1}$ and $B_{2}$ . The number of ways, in which these girls and boys can stand in a queue such that all the girls stand together, all the boys stand together, but $B_{1}$ and $B_{2}$ are not adjacent to each other, is :

A
144
B
72
C
96
D
120

Solution

Treat the girls block and the boys block as two super-units. They can be arranged in $2!$ orders.

The 3 girls inside their block can be permuted in $3!$ ways.

For the 4 boys: total arrangements without restriction are $4!$ , but we must subtract those where $B_{1}$ and $B_{2}$ stand adjacent. Treating $B_{1}, B_{2}$ as one unit gives $3!$ arrangements of three units multiplied by $2!$ internal orders of the pair, i.e. $3! \cdot 2!$ .

So valid boy arrangements: $4! - 3! \cdot 2! = 24 - 12 = 12$ .

Total arrangements: $2! \cdot 3! \cdot 12 = 2 \cdot 6 \cdot 12 = 144$ .

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JEE Main 2025 Jan 22 Shift 2, Mathematics Q2: Permutation

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