There are 9 children comprising of 3 bos and 6 girls. Let m denotes the number of ways when the 3 boys are in succession and n denotes the corresponding figure when the two end position are occupied by boys. If m = kn then k is equal to

Question

There are 9 children comprising of 3 bos and 6 girls. Let m denotes the number of ways when the 3 boys are in succession and n denotes the corresponding figure when the two end position are occupied by boys. If m = kn then k is equal to (A) 3 (B) 2 (C) 1 (D) (1/3)

Tardigrade Admin · Accepted Answer

m =3 ! ⋅ 7 ! B 1 B 2 B 3 G 1 G 2 ldots G 6 m = 3 C 2 2 ! 7 !=3 ! 7 !. Hence k =1

Q. There are 9 children comprising of 3 bos and 6 girls. Let $m$ denotes the number of ways when the 3 boys are in succession and $n$ denotes the corresponding figure when the two end position are occupied by boys. If $m = kn$ then $k$ is equal to

3

2

1

(1/3)

$m = 3! \cdot 7! B_{1} B_{2} B_{3} G_{1} G_{2} \dots G_{6}$
$m =^{3} C_{2} 2! 7! = 3! 7!$ . Hence $k = 1$

Q. There are 9 children comprising of 3 bos and 6 girls. Let m denotes the number of ways when the 3 boys are in succession and n denotes the corresponding figure when the two end position are occupied by boys. If m=kn then k is equal to

3

2

1

(1/3)

m=3!⋅7!B1​B2​B3​G1​G2​…G6​ m=3C2​2!7!=3!7!. Hence k=1

Q. There are 9 children comprising of 3 bos and 6 girls. Let $m$ denotes the number of ways when the 3 boys are in succession and $n$ denotes the corresponding figure when the two end position are occupied by boys. If $m = kn$ then $k$ is equal to

$m = 3! \cdot 7! B_{1} B_{2} B_{3} G_{1} G_{2} \dots G_{6}$
$m =^{3} C_{2} 2! 7! = 3! 7!$ . Hence $k = 1$