Question:
What are the number of ways to draw 4 balls from a box which contains
3 red balls and 2 yellow balls (without replacement)
My attempt:
You can either choose 3 red balls and 1 yellow ball OR 2 red balls and 2
yellow balls so the answer should be: $$3C3*2C1+3C2*3C2 = 5$$
However, tree diagram shows otherwise (number of ways to draw 4 balls = 10) so I must be wrong.
Thanks in advance
Best Answer
As you noted, there are $2$ ways to draw $4$ balls. Either draw $3$ red balls and $1$ yellows, or $2$ red balls and $2$ yellows.
In the first case, There are $\binom{4}{3}$ ways to order the sequence of $4$ balls.
In the first case, There are $\binom{4}{2}$ ways to order the sequence of $4$ balls.
Hence, the total is $\binom{4}{3}+\binom{4}{2}=4+6=\boxed{10}$