Question #194171

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) 


1
Expert's answer
2021-05-18T05:06:02-0400

1.

N=C52=5!3!2!=10N=C^2_5=\frac{5!}{3!2!}=10 ways


2.

N=C105C63=10!5!5!6!3!3!=5040N=C^5_{10}\cdot C^3_6=\frac{10!}{5!5!}\cdot \frac{6!}{3!3!}=5040 ways


3.

a.

N=C61C51C31=653=90N=C^1_6 \cdot C^1_5\cdot C^1_3=6\cdot5\cdot 3=90 ways


b.

N=C63=6!3!3!=20N=C^3_6=\frac{6!}{3!3!}=20 ways


c.

N=C52C31=5!2!3!3!2!=30N=C^2_5\cdot C^1_3=\frac{5!}{2!3!}\cdot \frac{3!}{2!}=30 ways


d.

N=C52C91=5!2!3!9!8!=90N=C^2_5\cdot C^1_9=\frac{5!}{2!3!}\cdot \frac{9!}{8!}=90 ways


e.

N=C113=11!8!3!=165N=C^3_{11}=\frac{11!}{8!3!}=165 ways


f.

N=C32C111+C33=3!2!11!10!+1=34N=C^2_3\cdot C^1_{11}+C^3_3=\frac{3!}{2!}\cdot \frac{11!}{10!}+1=34 ways


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