Question

In how many ways can 5 boys and 5 girls stand in a row so that no two girls may be together.

• A. $(5\mspace{6mu}!)^{2}$
• B. $5\mspace{6mu}!\mspace{6mu} \times 4\mspace{6mu}!$
• C. $5\mspace{6mu}!\mspace{6mu} \times 6\mspace{6mu}!$
• D. $6 \times 5\mspace{6mu}!$

Answer 1

Answer: (C)

$5\mspace{6mu}!\mspace{6mu} \times 6\mspace{6mu}!$

Answer 2

5 boys can be stand in a row $5\mspace{6mu}!$ ways

Now, two girls can't stand in a row together in $6P_{5}$ ways.

Total no. of required arrangement $= 5\mspace{6mu}!\mspace{6mu} \times^{6} ⥂ P_{5} = 5\mspace{6mu}!\mspace{6mu} \times 6\mspace{6mu}!$.