A counting problem

Probability Level 3

Consider three boxes , each containing $10$ balls labelled $1,2,\cdots ,10$ .Suppose one ball is 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 ?

120 160 150 130

2 solutions

Brian Charlesworth
Jan 29, 2018

Choose $3$ distinct integers from { $1,2,3,....,10$ } and assign the least of these to $n_{1}$ , the next least to $n_{2}$ and the greatest to $n_{3}$ . This can be done in $\dbinom{10}{3} = \boxed{120}$ ways and accounts for all the ways in which the balls can be chosen such that $n_{1} \lt n_{2} \lt n_{3}$ .

Jc 506881
Jan 28, 2018

Given $n_2$ , there are $n_2 - 1$ choices for $n_1$ and $10 - n_2$ choices for $n_3$ . So, the total number of ways the balls can be chosen such that $n_1 < n_2 < n_3$ is: $\sum_{n_2 = 1}^{10} (n_2 - 1)(10 - n_2) = 120$

A counting problem

2 solutions

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