Question

In how many ways can 5 boys and 3 girls be seated in a row so that each girl is between 2 boys?

Answer 1

Given: We have 5 boys and 3 girls 

To Find: Number of ways of seating so that 5 boys and 3 girls are seated in a row and each girl is between 2 boys 

The formula used: The number of permutations of n different objects taken r at a time 

(object does not repeat) is ⁿP_r =  \(\frac{n!}{(n-r)!}\)

The only arrangement possible is 

B__B__B__B__B 

Number of ways for boys =  ⁿP_r

= ⁵P₅

= \(\frac{5!}{(5-5)!}\) 

= \(\frac{5!}{0!}\)

= 120 

There are 3 girls, and they have 4 vacant positions 

Number of ways for girls = ⁴P₃ = 24 ways

= \(\frac{4!}{(4-3)!}\)  

= \(\frac{4!}{1!}\) 

= 24

Total number of ways = 24 × 120 = 2880

In 2880 ways 5 boys and 3 girls can be seated in a row so that each girl is between 2 boys.