4 Boys, 3 Girls Seated: Girls Together, B1 and B2 Not Adjacent

JEE Main 2025 — exact session not confidently identified. Previous Year Question.

Problem

4 boys and 3 girls are to be seated in a row such that all girls sit together and two particular boys, $B_1$ and $B_2$, are not adjacent to each other. The number of ways in which this arrangement can be done is:

Key insight. “All girls sit together” is handled by treating the three girls as a single block — but that block, once formed, is just one more object to arrange alongside the boys. The “$B_1$, $B_2$ not adjacent” condition is then a separate, second layer: instead of arranging $B_1$ and $B_2$ freely, they’re placed into two of the gaps left by the other objects, which automatically keeps them apart.

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Approach

Since all three girls must sit together, treat them as a single block. Setting $B_1$ and $B_2$ aside for a moment, that leaves the girls’ block plus the two other (unrestricted) boys — three objects in total, which can be arranged first. Arranging those three objects creates gaps between and around them; placing $B_1$ and $B_2$ into two different gaps guarantees they can never end up next to each other, since anything else seated between them would separate them anyway. Multiplying together the arrangements of the girls within their block, the arrangement of the three main objects, and the ways to place $B_1$/$B_2$ into non-adjacent gaps gives the total count.

Solution

Step 1 — Treat the girls as one block

The 3 girls form a single block, which can be internally arranged in $3! = 6$ ways.

Step 2 — Arrange the block with the two unrestricted boys

Excluding $B_1$ and $B_2$, there are 2 other boys plus the girls’ block — 3 objects total, arranged in $3! = 6$ ways.

Step 3 — Count the gaps available for $B_1$ and $B_2$

Arranging those 3 objects in a row creates 4 possible gaps (before the first object, between each pair, and after the last) where $B_1$ and $B_2$ could be inserted:

$$\_\ \times\ \_\ \times\ \_\ \times\ \_$$

Choosing 2 of these 4 gaps for $B_1$ and $B_2$ (one boy per chosen gap, so order matters since $B_1$ and $B_2$ are distinct) can be done in $4\times3=12$ ways — equivalently, ${}^4P_2 = 12$, which is the same as ${}^4C_2 \times 2! = 6\times2=12$.

Step 4 — Multiply all the pieces together

$$3! \text{ (girls within block)} \times 3! \text{ (arranging 3 objects)} \times {}^4P_2 \text{ (placing } B_1, B_2\text{)}$$

$$6\times 6\times 12 = 432$$

Answer

Common mistakes

• Arranging $B_1$ and $B_2$ as if they were part of the initial 3-object arrangement, rather than inserting them afterward into gaps — this either double-counts arrangements or fails to guarantee they’re non-adjacent.
• Using combinations instead of permutations when placing $B_1$ and $B_2$ into the gaps. Since $B_1$ and $B_2$ are distinct individuals, choosing gap A for $B_1$ and gap B for $B_2$ is different from the reverse — this needs ${}^4P_2=12$, not ${}^4C_2=6$.

Practise next

• 5 boys and 2 girls are to be seated in a row such that both girls sit together and two particular boys are not adjacent — count the arrangements using the same block-and-gaps method.