Selecting 4 members from groups and atleast 1 from group A 
If a team of four persons is to be selected from 8 males and 8
females, then in how many ways can the selections be made to include
at least one male.

I am able to find 3 ways to solve this questions:

*

*Total ways $-$ ways in which no male is included

*Make cases like 1 male 3 female, 2 male 2 females...

*Select 1 male and then fill rest three freely as

${8 \choose 1} * {15 \choose 3}$
I can't wrap my head around this. Rest 2 of them works but this method doesn't. I am missing something basic. Thanks.
 A: Methods 1 and 2 should work.

*

*${16\choose4}-{8\choose4}=1750$

*${8\choose1}{8\choose3}+{8\choose2}{8\choose2}+{8\choose3}{8\choose1}+{8\choose4}=1750$
Method 3 does not correctly count the set.  You would be counting M1 + M2W1W2 as a different arrangement than M2 + M1W1W2 even though they represent the same team.
A: The third method is wrong, you are overcounting,  because you can choose male A among the $8$ members, and male B while freely choosing from $15$ members, to create a team, and the same team can be created while choosing male B among the $8$ and male A among the 15, and the rest 2 members remaining the same.
The first two methods are correct though.
A: In method 3 you are not correctly counting the set. Some solutions would be included twice.
Let one boy be Ram another be Sam and the two woman be sita and rita.
Then one pair would become Ram+Sam x Sita x Rita and another pair would become Sam+Ram x Sita x Rita. Since like this many pairs can be counted twice your third method will not give you the correct answer.
A: You are overcounting scenarios where there are more than one male in the team when you solve it with the 3rd method.
Let's consider a particular combination where you select a male A first and then select
a male B and females X and Y.You could form the same team by selecting B first and then
selecting A, X and Y.
These two cases would be counted as different combinations while using the 3rd method
when in fact they are forming the same team.
By the time you get to counting teams in which all 4 members are male you would be counting 23 extra combinations for every team. ie:you would be counting 4!=24 cases for every team.
