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Consider three boxes, each containing 10 balls labelled $1,2,....,10$. Suppose one ball is randomly drawn from each of the boxes and is denoted by $n_i$, where $i$ represents the 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?

I tried doing this in different ways, but can't take the first step. I tried using multinomial theorem, but can't understand which terms to consider. Any help will be appreciated.

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closed as off-topic by user21820, Paul Frost, B. Goddard, Eevee Trainer, José Carlos Santos Mar 25 at 0:04

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Well if you take a ball in the middle, say it is $k$, then for the first one you have $k-1$ choises and for the third $10-k$ choises. So we have $$1\cdot 8 + 2\cdot 7 + 3\cdot 6 + 4\cdot 5 + 5\cdot 4 +6\cdot 3 + 7\cdot 2+8\cdot 1 = $$ $$= 2\cdot ( 1\cdot 8 + 2\cdot 7 + 3\cdot 6 + 4\cdot 5 )$$

$$= 2\cdot ( 8 + 14 + 18 + 20 )= 120$$

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  • $\begingroup$ Thanks a lot. I was trying to consider such cases, but the mistake I was doing was that I was trying to consider a ball from the first and last boxes. It didn't strike me to consider a ball from the middle box. $\endgroup$ – Arka Seth Mar 22 at 20:41
  • $\begingroup$ Ah extremely sorry for that. It is 2:45am here right now and in the midst of prepping for my sleep I'd forgotten about it. $\endgroup$ – Arka Seth Mar 22 at 21:12
  • $\begingroup$ No worry, thank you, have a nice sleep $\endgroup$ – Aqua Mar 22 at 21:12
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The number of ways to select the three balls is the same as the number of strictly ascending subsequences of length 3 taken from 1, 2, 3, ..., 10, which is the same as the number of subsets of size 3 taken from $\{1,2,3,\dots,10\}$, which is $$\binom{10}{3} = 120$$

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