In how many ways can 5 boys and 3 girls be seated in a row so that eac

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

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

Answer»

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.