Question

Consider three boxes, each containing $10$ balls labelled $1,2,\dots ,10.$ Suppose one ball is randomly drawn from each of the boxes.Denote by $n_i,$ the label of the ball drawn from the $i^{th}$ box, $(i=1,2,3).$ Then, the number of ways in which the balls can be chosen such that $n_1<n_2<n_3$ is :

• A. $82$
• B. $240$
• C. $164$
• D. $120$

Answer 1

Answer: (D)

$120$

Number of ways${=}^n{C}_3{=}^{10}{C}_3$

$=\frac{10!}{7!3!}$

$=\frac{10\times 9\times 8}{3\times 2}$

$=5\times 3\times 8=120$ ways.