Question

The number of ways in which $4$ men, $3$ boys, $2$ women can be seated in a row so that the men, the boys and the women are not seperated is

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

Answer 1

Answer: (C)

$4!$ $(3!{)}^2$ $2!$

The men can be taken as 1 block, women as 1, and boys as 1 block.
We have 3 block to rearrange

$M_1{M}_2{M}_3{M}_4$

$B_1{B}_2{B}_3$

$W_1{W}_2$

No. of permutations of 3 blocks = 3!

Additionally, No. of ways of rearranging the men = 4!
the boys = 3!
the women = 2!

Total $=3!\times (4!3!2!)$
$=4!\times (3!{)}^2\times 2!$