# Set theory problem

• Nov 27th 2012, 03:48 PM
wattskickin
Set theory problem
Need help solving the following. I cannot find enough information in the text to begin solving the problem.

1/4 of the five element subsets of (1,2,3...n) contain the element 7, determine n. n >5.
• Nov 27th 2012, 04:29 PM
HallsofIvy
Re: Set theory problem
There are $\begin{pmatrix}n \\ 5\end{pmatrix}= \frac{n!}{5! (n-5)!}$ subsets of {1, 2, 3, ..., n} that contain 5 numbers. 1/4 of that would be $\frac{n!}{5!(n-5)! 4}$ and "1/4 of the five element subsets of (1,2,3...n) contain the element 7" implies that must be an integer.
• Nov 27th 2012, 09:03 PM
wattskickin
Re: Set theory problem
I have been struggling with the equation for hours trying to solve for n without success. Is there enough information given to solve this problem?
• Nov 27th 2012, 09:55 PM
MarkFL
Re: Set theory problem
Yes, there is enough information given to solve for $n$.

HallsofIvy has told you that 1/4 of the number of subsets containing 5 elements is:

$N=\frac{n!}{4\cdot5!(n-5)!}$

Now, to find the number of subsets of cardinality 5 containing 7 as an element, we may use the fundamental counting principle to state:

$N={n-1 \choose 4}=\frac{(n-1)!}{4!((n-1)-4)!}=\frac{(n-1)!}{4!(n-5)!}$

Hence, we have:

$\frac{n!}{4\cdot5!(n-5)!}=\frac{(n-1)!}{4!(n-5)!}$

Now, you just need to solve for $n$.
• Nov 28th 2012, 08:00 AM
wattskickin
Re: Set theory problem
I think the equation I need to solve for n is

1/4*(n,5)=(n-1,4)

but I have no idea how to solve this equation
• Nov 28th 2012, 08:04 AM
Re: Set theory problem
This equation is the same as the last equation in MarkFL2 post. Multiply both sides by (n-5)!. Then divide both sides by (n-1)! (remember that n!/(n-1)! = n). Then multiply oth sides by 4!.
• Nov 28th 2012, 08:15 AM
Plato
Re: Set theory problem
Quote:

Originally Posted by wattskickin
I think the equation I need to solve for n is
1/4*(n,5)=(n-1,4) but I have no idea how to solve this equation

You have been told exactly what to do.

Solve $\frac{n!}{4\cdot5!(n-5)!}=\frac{(n-1)!}{4!(n-5)!}$