Question

Three boys of class X, four boys of class XI, and five boys of class XII sit in a row. The total number of ways in which these boys can sit so that all the boys of same class sit together is equal to

• A. $(3!{)}^2(4!\left)\right(5!)$
• B. $(3!)(4!{)}^2(5!)$
• C. $(3!)(4!)(5!)$
• D. $(3!)(4!)(5!{)}^2$

Answer 1

The correct option is A $(3!{)}^2(4!\left)\right(5!)$

We can think of three packets. One consisting of three boys of class $X$,
other consisting of $4$ boys of class $XI$ and last one consisting of $5$ boys
of class $XII$.

These packets can be arranged in $3!$ ways and contents of
these packets can be further arranged in $3!$ $4!$ and $5!$ ways, respectively.
Hence, the total number of ways is $3!\times 3!\times 4!\times 5!$