Question

In how many ways can $8$ men and $8$ women be arranged in a row so that they are positioned alternatively?

• A. $2\times 8{!}^2$
• B. $2\times 8!\times 9!$
• C. $8!\times 9!$
• D. $8{!}^2$

Answer 1

Answer: (A)

$2\times 8{!}^2$

$8$ mens , $8$ womens

Case - 1 : even places by men , odd places by women

Number of ways

$\Rightarrow 8!\times 8!\Rightarrow {(8!)}^2$

Case 2 even places by women ' odd places by men

number of ways

$\Rightarrow 8!\times 8!\Rightarrow {(8!)}^2$

Total number of ways

$\Rightarrow {(8!)}^2+{(8!)}^2\Rightarrow 2\times {(8!)}^2$

Option $A$