Ways to draw 3 balls A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?
My logic: total ways to draw 3 balls which is 9C3 - the case which does not have any black ball which is 6C2
So according to me answer is 
9C3 - 6C2 = 84 -15= 69
But answer is something else. Please help
 A: If a black ball must be included, the problem equals to draw $2$ balls from a box with $2$ black, $2$ white and $4$ red. So you have this six options:
Black Black.
White White.
Red Red.
Black White.
Black Red.
White Red.
A: Summarizing comments, your answer and approach is perfectly valid (and indeed the one I recommend as it saves time with computation), taking the number of outcomes overall without restriction, and subtracting away those outcomes where no blacks were drawn at all.
This gives a total of $$\binom{9}{3}-\binom{6}{3}=64$$
You could have done as the solutions author you are citing did and approached directly, breaking into cases based on the total number of blacks selected.  This is all well and good and arrives at the same answer, but requires noticeably more computation, especially if we were to extend the problem to asking about drawing at least one black ball when all of the other numbers in the problem were much larger.
Notice:
$$\binom{9}{3}=\binom{6}{3}\binom{3}{0}+\binom{6}{2}\binom{3}{1}+\binom{6}{1}\binom{3}{2}+\binom{6}{0}\binom{3}{3}$$
seen by Vandermonde's Identity and so
$$\binom{9}{3}-\binom{6}{3}=\binom{6}{2}\binom{3}{1}+\binom{6}{1}\binom{3}{2}+\binom{3}{3}$$
and so the answers are in fact the same, just written differently.
Your only mistake was either a typo or a lack of focus and using $\binom{6}{2}$ in place of $\binom{6}{3}$ in your attempt.
