A container has $20$ black balls, $15$ white balls and $18$ red balls. Each $2$ balls of the same color are identical. How many ways there are to extract balls from the container such that we choose at least $1$ black ball, $1$ white ball and $1$ red ball. No other limitations on the amount of balls extracted apply.
What I tried to do is apply inclusion/exclusion principle and count the number of possibilities where:
- $A_1$ = No black ball was extracted
- $A_2$ = No white ball was extracted
- $A_3$ = No red ball was extracted
Because each ball is either extracted or not, the total number of possibilities is $2^{53}.$ Applying the inclusion/exclusion we result in: $$|A_1| + |A_2| + |A_3| = 2^{33} + 2^{38} + 2^{35}$$ $$|A_1 \cap A_2| + |A_1 \cap A_3| + |A_2 \cap A_3| = 2^{15} + 2^{18} + 2^{20}$$ $$|A_1 \cap A_2 \cap A_3| = 1$$
The total number of possibilities is: $2^{53} - 2^{33} - 2^{38} - 2^{35} + 2^{15} + 2^{18} + 2^{20} - 1$
However the textbook's answer is: $20 \cdot 18 \cdot 15$. Just by looking at this, I have no idea how one can come up with an answer like this. Can someone please explain what am I doing wrong?
You are making things more difficult than necessary.
Concerning the black balls there are $20$ choices for the number of balls to be extracted: $1,2,3,4,\dots,20$.
Likewise there are $18$ choices for the red balls and $15$ for the white balls.
That gives $20\times18\times15$ choices in total.