There are 9 children comprising of 3 boys and 6 girls. Let m denotes the number of ways they can be seated in a line when all 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 boys and 6 girls. Let m denotes the number of ways they can be seated in a line when all 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! <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/58ed18013877262ce7273f1fd8fe4e57-.png" /> n = 3 C 2 ⋅ 2 ! 7 !=3 ! 7 ! Hence k = 1

Q. There are 9 children comprising of 3 boys and 6 girls. Let m denotes the number of ways they can be seated in a line when all 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

$m = 3! \cdot 7!$

$n =^{3} C_{2} \cdot 2! 7! = 3! 7!$
Hence $k = 1$

Q. There are 9 children comprising of 3 boys and 6 girls. Let m denotes the number of ways they can be seated in a line when all 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! n=3C2​⋅2!7!=3!7! Hence k=1

Q. There are 9 children comprising of 3 boys and 6 girls. Let m denotes the number of ways they can be seated in a line when all 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

$m = 3! \cdot 7!$

$n =^{3} C_{2} \cdot 2! 7! = 3! 7!$
Hence $k = 1$