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Number of ways to select three items from $n$ distinct items are :- $\binom{n}{3}$

Modified version of the problem :- Given $n$ boxes, each box having a particular amount of items , what is the number of ways to select three items such that each of the three items belong to different boxes ?

Example:- Let $n=3$,

First box contains $x1=2$ items.

Second box contains $x2=3$ items.

Third box contains $x3=1$ item(s).

Hence, total number of ways to select three items :- $6$(ways).

For any given $n$ and given $x1,x2,x3.......$till $n^th$ index, what are the number of ways to select three items such that each item belongs to a different box ?

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Well, first you'd have to choose three boxes to pull from, and then you'd have to pull one item from each box.

Here is a general way to come up with all the combinations:

$\displaystyle \large \sum_{a=1}^{n-2}\sum_{b=a+1}^{n-1}\sum_{c=b+1}^{n}x_ax_bx_c$.

The triple summation is something constructed to traverse through each possible combination of three boxes without repetition, while $x_ax_bx_c$ multiplies the number of items in the three boxes for each combination.

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  • $\begingroup$ So its basically the sum of all the product-triplets in the array of boxes $\endgroup$ Commented Feb 8, 2020 at 20:48
  • $\begingroup$ @ShaliniTomar exactly $\endgroup$ Commented Feb 8, 2020 at 20:48

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