There are 15 aliens: 8 Plutonians and 7 Venusians. How many ways can I form a group of 4 with at least 1 Venusian?

Question

I understand that this answer is $1295$. I can get this by either of:

$${{7}\choose{1}}{{8}\choose{3}}+ {{7}\choose{2}}{{8}\choose{2}}+{{7}\choose{3}}{{8}\choose{1}}+{{7}\choose{4}}{{8}\choose{0}}$$ or $${{15}\choose{4}}-{{7}\choose{0}}{{8}\choose{4}}.$$

But what cases am I double counting if I instead think like this: Choose the one Venusian ${{7}\choose{1}}$ then choose any three others ${{14}\choose{3}}$.

i.e. Why is ${{7}\choose{1}}{{14}\choose{3}}$ not logically correct? I know I am double counting but cannot tell how.

Answer 1

If things like this are not clear then it is often handsome to look at a similar situation with extra small numbers.

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Form by brute force a group of $2$ out of $1$ Plutonian and $2$ Venusians with at least one Venusian.

Applying your method we get $\{V_1,P_1\},\{V_1,V_2\},\{V_2,P_1\},\{V_2,V_1\}$.

But this with:$\{V_1,V_2\}=\{V_2,V_1\}$

Answer 2

If the first Venusian was to be group leader, your method would work, but this isn't the case, and you have double counted $v_1v_2p_ip_j$, triple counted with groups such as $v_1v_2v_3p_i$ and quadruple counted with groups such as $v_1v_2v_3v_4$ (i.e. this group is listed $4$ times).