Let m denotes the number of ways in which 5 boys and 5 girls can be arranged in a line alternately and n denotes the number of of ways in which 5 boys and 5 girls can be arranged in a circle so that no two boys are together. If m = kn then the value of k is

Question

Let m denotes the number of ways in which 5 boys and 5 girls can be arranged in a line alternately and n denotes the number of of ways in which 5 boys and 5 girls can be arranged in a circle so that no two boys are together. If m = kn then the value of k is (A) 30 (B) 5 (C) 6 (D) 10

Tardigrade Admin · Accepted Answer

m = 5! · 2 · 5! = 10 · 4! · 5! n = 4! · 5! Hence m =10 n ⇒ k =10

Q. Let m denotes the number of ways in which 5 boys and 5 girls can be arranged in a line alternately and n denotes the number of of ways in which 5 boys and 5 girls can be arranged in a circle so that no two boys are together. If m = kn then the value of k is

30

5

6

10

$m = 5! \cdot 2 \cdot 5! = 10 \cdot 4! \cdot 5!$
$n = 4! \cdot 5!$
Hence $m = 10 n \Rightarrow k = 10$

Q. Let m denotes the number of ways in which 5 boys and 5 girls can be arranged in a line alternately and n denotes the number of of ways in which 5 boys and 5 girls can be arranged in a circle so that no two boys are together. If m = kn then the value of k is

30

5

6

10

m=5!⋅2⋅5!=10⋅4!⋅5! n=4!⋅5! Hence m=10n⇒k=10

$m = 5! \cdot 2 \cdot 5! = 10 \cdot 4! \cdot 5!$
$n = 4! \cdot 5!$
Hence $m = 10 n \Rightarrow k = 10$