A box contains $6$ white balls and $5$ black balls. Find the number of ways $4$ balls can be drawn from the box if two must be white?
1st Approach I thought was
Selecting two white balls from the bag $^6C_2$ and then selecting the other two from the remaining balls in the bag $^9C_2$, so the answer should be $^6C_2\cdot{^9}C_2$.
2nd Approach was
Making the cases for $2$ white balls then $3$ white balls and then $4$ white balls and then adding up all the cases to get the answer.
The problem is the answer from both these approaches is different, so I want to know which one is correct and why? and which one is wrong and why? because only one can be correct.
Best Answer
Your first approach is incorrect. It assumes the white balls are distinct, but it overcounts the cases of more than two white balls. You would count the case of $W1,W2,W3,B1$ three times, once with the first two white balls $W1, W2$, once with $W1,W3$ and once with $W2,W3$. In each case the third white ball is one of the two remaining balls.