Answer to Question #194171 in Discrete Mathematics for Saoirse lei kjelbe

Question

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) 

Answer 1

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