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

Question

Consider three boxes, each containing 10 balls labelled 1, 2, ..., 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. Then the number of ways in which the balls can be chosen such that $n_1<n_2<n_3$ is:

Accepted Answer

Choose 3 distinct labels from $\{1,\ldots,10\}$; there's exactly one way to assign them in increasing order:

$$\binom{10}{3} = \frac{10\times9\times8}{6} = 120$$

Solutions · 1