By definition, we have
[tex]C(n,k) = \displaystyle \binom{n}{k}=\dfrac{n!}{k!(n-k)!}[/tex]
Where
[tex]n! = n(n-1)(n-2)(n-3)\ldots 3\cdot 2[/tex]
So, in your case, we have
[tex]C(7,4) = \displaystyle \binom{7}{4}=\dfrac{7!}{4!3!}=\dfrac{7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2}{(4\cdot 3\cdot 2)\cdot (3\cdot 2)}=7\cdot 5=35[/tex]
