Binomial coefficient (7 choose 3)=Binomial coefficient (7 choose x)

Question

How can I solve the following equation:

$$\frac{7!}{3!(7-3)!}=\frac{7!}{x!(7-x)!}$$

What seems problematic to me in particular is the term $$x!(7-x)!$$.

Answer 1

What if your choice of $x$ flipped the values of the factorials?

Answer 2

you have from the denominator that $$x!(7-x)! = 3! 4! = 4! 3!$$ Choose $x = 4$.

Answer 3

You can write your equation like $\binom{7}{3}=\binom{7}{x}$.

Then you can use the known rule: $\binom{x}{p}=\binom{x}{x-p}$ to conclude that

$$x=3 \lor x=7-3=4$$

are the solutions to your problem.

Answer 4

In the spirit of the other two answers: try to prove that for all $\;n,\,k\in\Bbb N\cup\{0\}\;,\;\;\;k\le n\;$ , we have that

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

Answer 5

You have to solve $$x!(7-x)! = 3!4!$$

Note that $5$ does not divide $3!4!$. Thus $x<5$ and $7-x < 5 \iff x > 2$.

So $x=3$ and $x=4$ are the solutions.

Answer 6

Use the equation

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

to get

$$\binom{7}{3}=\binom{7}{4}.$$

To see that $3$ and $4$ are the only possible solutions, take a look at Pascal's triangle and notice the behavior of the binomial coefficients.

(This is not rigorous but Pascal's triangle + thinking about the meaning of $\binom{n}{k}$ should give you the intuitive idea why 3 and 4 are the only things that work.)

Hope this helps, comment if you'd like me to elaborate further.