# Finding number of subsets in a set

I have two questions that I've been working on.

Given: Set A = { a, b, c, d, e, f, g, h }

1. How many sets of size 5 does A have which contain the element b?

I think I have this one. My solution is 8C5 - 7C5 = 35 sets (Please let me know if this is wrong!).

Here is the second one that throws me off:

1. How many sets of size 5 does A have which contain the element c but do not contain e?

I am not sure where to start. the "do not contain e" part throws my brain off. Any guidance would be helpful!

• "the "do not contain e" part throws my brain off" Why? When it comes to picking elements just... don't pick $e$. If you want you can treat it as the $e$ wasn't an element in the set to begin with. How many subsets of $\{a,b,c,d,f,g,h\}$ have five elements and contain $c$. – fleablood Oct 15 '19 at 2:39

1) Another way to do 1) is that you must have $$b$$ and there are four other elements from the remaining $$7$$ elements to pick. So there are $${7\choose 4} = 35$$.

2) So you must have a $$c$$ and you must have $$4$$ other elements but none of them can be $$e$$. So the other $$4$$ can be any of $$a,b,d,f,g,h$$. So there are $${6 \choose 4}$$ ways to do this.

Or you can do inclusion exclusion.

$${8\choose 5}$$ are the total.

$${7\choose 5}$$ are those that don't contain $$c$$ so $${8\choose 5} - {7\choose 5}$$ are those that do contain $$c$$.

And $${8\choose 5} - {7\choose 5}$$ are those that do contain $$b$$ so $${8\choose 5} - ({8\choose 5} - {7\choose 5})$$ are those that don't contain $$b$$. (Um... are you being to see we doing things by taking the compliment of things that don't have a property can be .... redundant?)

And then those that do contain $$c$$ but not $$b$$ are .... Hmm, okay. I don't know how to do this if compliment of those that do/don't is your only tool.

A subset of size $$5$$ that contains $$c$$ must have $$4$$ elements other than $$c$$. There are $$7$$ elements other than $$c$$ in $$A$$, but we're not allowed to choose $$e$$ so we have only $$6$$ to choose from. That gives $${6\choose4}=15$$. By $${6\choose4}$$ I mean the same thing as $$_6C_4$$.