You should use here the formula for combinations
[tex]C_n^k=\dfrac{n!}{k!(n-k)!}.[/tex]
In your case n=7, k=4, then
[tex]C_7^4=\dfrac{7!}{4!(7-4)!}=\dfrac{7!}{4!\cdot 3!}=\dfrac{4!\cdot 5\cdot 6\cdot 7}{4!\cdot 2\cdot 3}=5\cdot 7=35.[/tex]
Answer: 35 ways, option A.
