Permutations & Combinations
Permutation Combination
nta_pyq_2025_jan
Grade 11
Question:
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:
Step-by-Step Solution
Key Concept: Treat all girls as one block and all boys as one block: $2!$ ways to order blocks; $3!$ inside girls. Inside boys, count arrangements of $4$ where $B_{1},B_{2}$ are \emph{not} adjacent: $4!-(3!\cdot 2!)$.
Block ordering (girls-then-boys vs boys-then-girls): $2!=2.$
Inside girls: $3!=6.$
Inside boys, $B_{1}B_{2}$ not adjacent: total $4!$ minus arrangements treating $B_{1}B_{2}$ as a single block ($3!\cdot 2!$):
$24-12=12.$
Answer: $2\cdot 6\cdot 12=144.$
Correct Answer: 2