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  4. COMBINATIONS (24 pts) 1. Patrick has assignments in 5 subjects. He can only do two …
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COMBINATIONS (24 pts) 1. Patrick has assignments in 5 subjects. He can only do two assignments. In how many ways can he do two assignments? (3 pts) 2. In how many ways can a group of 5 men and 3 women be made out of a total of 10 men and 6 women? (3 pts) 3. A box contains 6 red, 5 blue and 3 white balls. In how many ways can we select 3 balls such that a. They are of different colors? (3 pts) b. They are all red? (3 pts) c. Two are blue and one is white? (3 pts) d. Exactly 2 are blue? (3 pts) e. None is white? (3 pts) f. At least two are white? (3 pts)

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Here's the Solution to this Question

1.

N=C^2_5=\frac{5!}{3!2!}=10 ways


2.

N=C^5_{10}\cdot C^3_6=\frac{10!}{5!5!}\cdot \frac{6!}{3!3!}=5040 ways


3.

a.

N=C^1_6 \cdot C^1_5\cdot C^1_3=6\cdot5\cdot 3=90 ways


b.

N=C^3_6=\frac{6!}{3!3!}=20 ways


c.

N=C^2_5\cdot C^1_3=\frac{5!}{2!3!}\cdot \frac{3!}{2!}=30 ways


d.

N=C^2_5\cdot C^1_9=\frac{5!}{2!3!}\cdot \frac{9!}{8!}=90 ways


e.

N=C^3_{11}=\frac{11!}{8!3!}=165 ways


f.

N=C^2_3\cdot C^1_{11}+C^3_3=\frac{3!}{2!}\cdot \frac{11!}{10!}+1=34 ways


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Question ID: mtid-5-stid-8-sqid-2990-qpid-1689