Consider three boxes, each containing 10 balls labelled 1, 2, . ,...

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 ${=}^{10}{C}_{3} . 1{=}^{10}{C}_{3} = \dfrac{\left(10 !\right)}{\left(3 !\right)\left(7 !\right)} = \dfrac{10 \times 9 \times 8}{3 \times 2} = 120$.