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I am a biologist and not a real mathematician. Hence some of the answers featured here are sometimes too complicated.

My question is:

I have set of 8 genes named PBX1,ESX1,PIM1,HBB,HBG,BCL11A,KLF4,GATA2

First I wanted to know how many different combinations of 6 I could make without caring about the order. I know now that this can be explained by:

(8x7x6x5x4x3)/(6x5x4x3x2x1) = 28 different combinations

However, the second problem is a bit different. I want to know within these sets of 6 genes how frequently subsets of 4 or 3 or 2 genes are represented (again without caring about the order).

for example: how often does the combination (PBX1,ESX1,PIM1,HBB) occur within these larger sets of 6.

I hope someone is able to help. Please consider my lack of real mathematical knowledge in your answering.

Thanks in advance!!

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So consider a fixed 4 tuple of genes, now we need to choose 2 more from the remaining 4 to make a six. So $\binom{4}{2}=6$ possible ways. So then out of the 28 different combinations, we have 6 of them containing the same 4-tuples.

By a similar argument, for a fixed three genes, we have $\binom{5}{3}=10$ containing the same three tuples.

And then for a fixed two genes, we have $\binom{6}{4}=15$ containing the same two tuples!!

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  • $\begingroup$ Thank you for your answer! That really helps! $\endgroup$ – pr94 Apr 17 at 15:21
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The labels just make things look more complicated than they are. Say the genes are $(g_1, g_2, \cdots, g_8)$. Then, as you say, the number of ways to choose an ordered collection of $6$ of these is $$\binom 86=28$$

Having singled out, say, $g_1, g_2, g_3, g_4$ and insisting that that these be part of your collection, we now just have to choose $2$ from $g_5, g_6, g_7, g_8$. The number of ways to do that is $$\binom 42=6$$

In general, if you select $i≤6$ genes and require them to be in your collection of $6$, then you have to choose $6-i$ from the remaining $8-i$ so the answer in that case would be $$\binom {8-i}{6-i}$$

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