The number $\binom{8}{4}$ is equal to the number of subsets of size 4 of the set $\{1, \dots, 8\}$ I was asked to proof if is true and give a counter example if it is false.
However I prefer True.
since all the numbers 1-8 insides the brackets are in the sets.
I'm I correct?
 A: Here is an intuitive proof:
For the first element in the subset, we have $8$ choices of which element to include. Because sets do not repeat elements, for the second element, we have $(8-1)$ choices, and for the third $(8-2)$, and for the fourth, $(8-3)$. This gives us the value $\frac{8!}{(8-4)!}$ as the number of possible subsets. However, because sets are unordered, we overcounted. Because each subset is of four elements, and the number of permutations of four elemetns is $4!$, we have $4!$ times the true number of subsets. Thus, the final number is $\frac{8!}{4!(8-4)!}$, which is by definition $\binom{8}{4}$.
A: The claim is true.
We know that $$ \binom84 = \frac{8\times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} $$
$8\times 7 \times 6 \times 5$ is the number of ways to pick 4 distinct numbers out of 8 distinct numbers one by one, because you have 8 choices when you pick the first number, then 7 choices when picking the second, and so on.
In this way, each combination of 4 numbers is counted $4 \times 3 \times 2 \times 1$ times, because there are $4 \times 3 \times 2 \times 1$ ways to permute 4 distinct numbers.
Hence the result.
