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 $i^{\text{th}}$ 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. $120$
• B. $164$
• C. $82$
• D. $240$

Answer 1

The correct option is A $120$

$n=10$
Select any three distinct numbers by ${}^{10}{C}_3$ ways.
Then assign the biggest number to $n_3$, the second biggest to $n_2$ and the smallest to $n_3.$
$\therefore$ Required number of ways in which the balls can be chosen
$={}^{10}{C}_3=\frac{10!}{3!7!}=\frac{10\times 9\times 8}{3\times 2}=120$