JEE Main 2026 — Apr 4, Shift 1 — Mathematics solution

JEE Main2026Apr 4, Shift 1Mathematics

The number of ways, of forming a queue of $4$ boys and $3$ girls such that all the girls are not together, is:

A
$5040$
B
$3050$
C
$3410$
D
$4320$

Solution

Total arrangements of $7$ distinct people: $7! = 5040$ .

Arrangements where all $3$ girls stand together: treat the girls as one block, giving $5$ entities to arrange ( $4$ boys plus the block) in $5!$ ways, with $3!$ internal orderings: $5! \cdot 3! = 120 \cdot 6 = 720$ .

Required count $= 7! - 5! \cdot 3! = 5040 - 720 = 4320.$

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