# Combinations balls and three boxes [closed]

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 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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## 2 Answers

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 $$1\cdot 8 + 2\cdot 7 + 3\cdot 6 + 4\cdot 5 + 5\cdot 4 +6\cdot 3 + 7\cdot 2+8\cdot 1 =$$ $$= 2\cdot ( 1\cdot 8 + 2\cdot 7 + 3\cdot 6 + 4\cdot 5 )$$

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

• 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. – Arka Seth Mar 22 at 20:41
• 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. – Arka Seth Mar 22 at 21:12
• No worry, thank you, have a nice sleep – Aqua Mar 22 at 21:12

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 $$\binom{10}{3} = 120$$