Consider $3$ boxes,each containing $10$ balls,labelled $1,2,...10$. 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?

  • $\begingroup$ $\sum_{i=3}^{10}\sum_{j=2}^{i-1}\sum_{k=1}^{j-1}1$. This summation is the answer and think about its meaning. $\endgroup$ – mastrok Sep 7 '15 at 5:47

HINT: Say that a draw is good if $n_1<n_2<n_3$. Each good draw produces a subset $\{n_1,n_2,n_3\}$ of $\{1,2,\ldots,10\}$ with exactly $3$ members. Suppose that $T$ is any set of $3$ distinct members of $\{1,2,\ldots,10\}$.

  • How many good draws result in the subset $T$?
  • How many such subsets $T$ are there?

A neat way to do it is, consider all combinations which $n_1\neq n_2 \neq n_3$, this is easily counted by $10\times 9\times 8 =720$.

Now think of this, for all choices of $n_1\neq n_2 \neq n_3$, there is only one ordering of these 3 numbers such that $n_1\le n_2 \le n_3$. There are $3\times 2\times 1 =6$ orderings for each choice of $n_1,n_2,n_3$, thus the required number of combination is $720 \div6=120$.

  • $\begingroup$ expanding on your approach I came up with the following solution,its a brute force method,I would love to hear your input.So I am trying to determine all possible triplets Lets start with the first possible combination ,i.e (1,2, _) Now _ can be filled in 8 ways,again the next combination (1,3, _) where _ can be filled in 7 ways ,so going on like this we get 8+7+6+...+1=36 ways for (1, _, _) Now for (2, _, _) we get 7+6+...1=28 ways Same way,we get 36+28+21+15+10+6+3+1=120 ways in total. (the last case being (8,9,10). So,is this correct? $\endgroup$ – bandit_king28 Sep 7 '15 at 7:13
  • $\begingroup$ Great, your direct counting is absolutely correct and most straight forward. $\endgroup$ – mastrok Sep 7 '15 at 7:22

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