# What is this set theory question even asking me?

I saw some other questions that had similar titles and checked them out, but none of them seemed to match this format. The question I have is:

One quarter of the five-element subsets of $\{1, 2, 3, \cdots, n\}$ contain the element $7$. Determine $n (\ge 5)$.

I don't understand what this question wants me to do. I understand what a subset is, and I know what elements are, but I don't know what "Determine $n (\ge 5)$" is asking me to do.

-
"What value must $n$ be in order for exactly one quarter of the five-element subsets of $\{1, 2, 3, \cdots, n\}$ to contain 7? $n$ must be $\ge 5$." –  Matthew Jan 28 '13 at 19:01
@Matthew That's exactly the information I was looking for. If you had posted this as an answer instead of a comment I'd be marking this as answered by you. Thanks. –  Dave Jan 28 '13 at 19:04
Glad I could help :) I am very much a mathematics neophyte, so it's probably better for me to have a low reputation anyway :P –  Matthew Jan 28 '13 at 22:26

## 2 Answers

Hint: There are $\dbinom{n}{5}$ five-element subsets.

How many five-element subsets are there that contain the number $7$? We have to choose $4$ elements from the remaining $n-1$ to keep company with the $7$.

For what value(s) of $n\ge 5$ is the second number one-quarter of the first?

Remark: The insistence that $n\ge 5$ is a bit picky, but in principle necessary. Suppose for example that $n=4$. There are $0$ ways to choose a five-element subset of $\{1,2,3,4\}$. There are also $0$ ways to choose a five-element subset that contains $7$. And $\frac{0}{4}=0$!

-
Thanks to Matthew who commented above I know what it's asking me to do. You're answer, though very helpful, didn't really answer what I was supposed to be doing, but guided me towards the first steps to find that answer. I certainly appreciate the response, though! –  Dave Jan 28 '13 at 19:07
Get an expression for the number of five-element subsets that contain $7$. I hinted on how to do that. Set that equal to $1/4$ of $\binom{n}{5}$. You get an equation in $n$. Solve. Want to leave the rest to you, unless you run into trouble. –  André Nicolas Jan 28 '13 at 19:10
@Dave: I think you might misunderstand, as I believe you will answer "must be $\ge 5$". This is not the answer, just an obvious necessary condition. AN guided you in the eight direction: you can calculate a value $n$ using the above: $\frac{1}{4} = \frac{\binom{n-1}{4}}{\binom{n}{5}$, which is what Andre pointed our. I tried finding a clever way of doing this, failed, and by trial and error it's at least 10; but one will work as your teacher asked you to determine it. –  gnometorule Jan 28 '13 at 19:19
Sorry for typos which I cannot correct on my phone, and won't be home in 5 mins...if a moderator can add the closing dollar, appreciated. –  gnometorule Jan 28 '13 at 19:21
@gnometorule: Don't need clever, one just expands, almost everything cancels. But if you feel like slightly clever, use identity $\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1}$. –  André Nicolas Jan 28 '13 at 19:26

"What value must n be in order for exactly one quarter of the five-element subsets of {1,2,3,⋯,n} to contain 7? n must be ≥5." Credit goes to @Matthew in the first comment on my question.

-