Question

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

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

Answer 1

The correct option is C $(4!)(3!{)}^2(2!)$

Consider $4$ men as one unit, internal arrangement of four men is $4!$.
Consider $3$ boys as one unit, internal arrangement of three boys is $3!$.
Consider $2$ women as one unit, internal arrangement of two women is $2!$
Now $4m+3b+2w=3$ objects, number of arrangements $=3!$
So, total number of ways $=3!\times 4!3!2!=4!(3!{)}^{2}2!$