Marble Distribution for Maria
Liz has 11 marbles of different colors. In how many ways can she choose 3 of the marbles, if she refuses to choose the green one and the purple one at the same time?
- 1
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- 9
- $\frac{a}{b}$
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- 0
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- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
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- $\cap$
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- $\infty$
Solution
We can divide this into cases:
Case 1: Liz chooses the green marble, and not the purple marble. In this case, Liz must choose 2 more marbles from the remaining 9 marbles (since she won't choose the purple marble). So, there are $ \binom{9}{2}=36 $ choices in this case.
Case 2: Liz chooses the purple marble, and not the green marble. As in Case 1, Liz must choose 2 more marbles from the remaining 9 marbles. So, there are $ \binom{9}{2} $ choices in this case.
Case 3: Liz chooses neither the green marble nor the purple marble. In this case, Liz must choose three marbles from the nine remaining marbles, and there are $ \binom{9}{3}=84 $ choices in this case.
The total number of possible choices is $ 36+36+84=\boxed{156} $.
Alternatively, there are $ \binom{11}{3}=165 $ ways to select three marbles. Of those, $ \binom{9}{1}=9 $ ways contain both the purple marble and the green marble. Therefore, there are $ 165-9=\boxed{156} $ ways to select three marbles such that purple and green marbles are not both chosen.
Freeflows