# In combinatorics, is ${13 \choose 5}$ the same as ${13 \choose 8}$?

So I got a question in college on combinatorics, I am supposed to find a coefficient. So I do the math according to the binomial theorem and I ended up with the answer ${13 \choose 5}$. But my teacher has written down the answer as ${13 \choose 8}$. When I calculate them I notice they are in fact the same number. So is it okay I answer ${13 \choose 5}$ when the right answer is ${13 \choose 8}$ when searching for a coefficient?

Forgive me if this is a silly question I am new to combinatorics.

• They are equal, so both are (the same) valid answer. You could just write $1287$ or even $10\cdot 2^7+7$ and it would still be correct. Sep 19, 2016 at 18:40
• To choose $5$ objects out of $13$ is the same as to choose $8$ (think of putting the $5$ in an urn, then put the remaining $13-5=8$ into a second urn). In general $\binom nk=\binom n{n-k}$ for the same reason.
– lulu
Sep 19, 2016 at 18:41
• Sorry but I don't grasp that. How can choosing 5 objects out of 13 be the same as choosing 8? Makes no sense in my head... Sep 19, 2016 at 18:42
• Intuitively, select $5$ elements from $13$ elements is the same of don't choose the remaining $8$ elements from $13$. Sep 19, 2016 at 18:43

Yes, they are the same!

Why? Well, $\binom{13}{8}$ is the number of ways to select $8$ people to be in a committee, from a group of $13$ people. But we could just as well choose $5$ people to not be on the committee. Choosing $8$ people to be on the committee is the same as choosing $5$ people to leave out. So $\binom{13}{8} = \binom{13}{5}$.

In general, it is a fact that $$\binom{n}{k} = \binom{n}{n-k},$$ and this is true for the same reason as I described.

$${n \choose k}=\frac{n!}{k!(n-k)!}=\frac{n!}{(n-k)!k!}=\frac{n!}{(n-k)!(n-(n-k))!}= {n\choose n-k}$$

$13 \choose 5$ is the number of ways to select five items out of $13$. If you take five items out, you leave eight behind, so the number of ways to select five and eight are equal. This is general. If you want to choose $k$ items out of $n$, we have ${n \choose k}=\frac {n!}{k!(n-k)!}={n \choose n-k}$

It's not a silly question at all. Those two are, in fact, the same, and what you've stumbled upon is a more general pattern. There are mainly two ways of seing this.

The formula:
We have $\binom{13}{8} = \frac{13!}{8!\cdot 5!}$ and $\binom{13}{5} = \frac{13!}{5!\cdot 8!}$. They look very similar to me.

The application:
If you have $13$ balls in a box, then $\binom{13}{5}$ is the number of ways to pick out $5$ of them. However, you could just as well have chosen $8$ of them to leave in the box, and say you'd pick out whichever balls are left. That's the same thing, and thus the number of ways to do each of them should be the same.

The general statement:
We have, for any natural numbers $n \geq r$ that $\binom{n}{r} = \binom{n}{n-r}$. In your case, we have $n = 13$, $r = 8$ and $n-r = 5$. But it also tells us, for instance, that $\binom{1000}{597} = \binom{1000}{403}$, or any other such relation.