JEE Main 2019 — Permutation Combination Question with Solution

Each box contains $10$ balls numbered from $1$ to $10$.

$n_1,n_2,n_3$ are numbers on the balls drawn from the box $B_1,B_2$ and $B_3$ respectively such that $n_1<n_2<n_3$.

i.e., all $3$ numbers $n_1,n_2,n_3$ must be different and can be arranged only in one way (increasing).

Now $n_1,n_2,n_3$ can be selected in ${}^{10}{C}_3$ ways.

Hence, total number of ways ${=}^{10}{C}_3.1{=}^{10}{C}_3=\frac{\left(10!\right)}{\left(3!\right)\left(7!\right)}=\frac{10\times 9\times 8}{3\times 2}=120$.