# Finding $\binom{n}{4}$ if $\binom{n}{6}=\binom{n}{9}$

I was trying to solve a question related to combinatorics as follows :

A group contains $n$ persons. If the number of ways of selecting $6$ persons is equal to the number of ways of selecting $9$ persons, then find the number of ways of selecting $4$ persons from the group.

I dont have a step by step solution for concluding that $n=15$.

I tried to solve for $n$ but I am stuck at $(n-9)(n-8)(n-7)=(1/504)$

• $(15-6)6=(15-9)9$ now use stirling number Aug 24, 2018 at 13:22
• sorry , i have just corrected my question , i just dont know how they get n=15 from given equality Aug 24, 2018 at 13:25
• The number $\binom{n}{4}$ can be either $0$, $1$, $5$ or $1365$. Aug 24, 2018 at 13:35
• If you use the formula $\binom n k=\frac{n!}{k!(n-k)!}$ you should get $(n-6)(n-7)(n-8)=504$, not the equation you gave. Aug 24, 2018 at 13:41

Remember the reflection/symmetry identity:

$$\binom{n}{k} = \binom{n}{n-k}$$

Notice that the lower indices sum to $n$. Thus, $n = 6+9 = 15.$

The rest follows.

• ahh, this answer is much better than my one. well done! Aug 24, 2018 at 13:53

If $n<6$ then the number of ways to select $6$ people is the same as the number of ways to select $9$ people; in both cases this is $0$. For $n<6$ we can easily compute $\binom{n}{4}$: $$\binom{5}{4}=5,\qquad\binom{4}{4}=1,$$ and $\binom{n}{4}=0$ when $n<4$.

If $n>6$, then by selecting $6$ people we leave $n-6$ people out. So the number of ways to select $6$ people is the same as the number of ways to select $n-6$ people. This suggests that $n-6=9$ and so $n=15$, and $\binom{15}{4}=1365$.

• A good answer that catches a case the poser clearly did not intend. Aug 24, 2018 at 14:33

Both the other answers explain why $n=15$ is possible. To show that there are no other possibilities (except the degenerate case $n<6$ which Servaes points out also works), note that if $1\leq r\leq n/2$ then $\binom nr=\frac{n-r+1}{r}\binom n{r-1}>\binom n{r-1}$, so for fixed $n$, $\binom nr$ is increasing with $r$ for $r\in\{0,n/2\}$ and similarly is decreasing for $r\in\{n/2,n\}$.

Thus for fixed $n$ and $k>0$, $\binom nr=k$ has at most two solutions, meaning that there are no nontrivial solutions to $\binom nr=\binom ns$ except for $s=n-r$.