Thread: Combination

I have a few problems I was wondering I was doing them right. Here they are:

1) In how many ways can 5 starting positions on a basketball team be filled with 10 men whom can play any position.

C(10,5) = 252

2) Nine people are going on a skiing trip in 3 cars that can hold 2, 4, and 5 passengers, respectively. In how many ways is it possible to transport nine people to the ski logde using all cars?

P(2,2)*P(4,4)*P(5,3) = 2*24*60 = 2880

3) The Detroit Music Company has purchased the rights to 10 different songs and it plans to release a new cd with 5 of them.
a) How many different cds are possible if the order of songs is important?
b) not important?

a)P(10,5) = 30,240
b)C(10,5) = 252

4) From a group of 4 men and 5 women, how many committees of size 3 are possible
a) with no restrictions?
b) with 1 man and 2 women?
c) with 2 men and 1 woman if a certain man must be on the committee?

a)C(9,6) = 84
b)C(4,1) * C(5,2) = 4 * 10 = 80
c)P(4,1) * C(3,1) * C(5,1) = 4 * 3* 5 = 60

5) Wood County sheriff classifies crimes by age (in years) of criminal and whether the crime is violent or nonviolent. As shown below, a total of 150 crimes were reported by the sheriff last year.
Under 20
20 to 40
40 or older
Total
Violent
27
41
14
82
Nonviolent
12
34
22
68
Total
39
75
36
150


a)What is the probability of selecting a case to analyze and finding it involved a violent crime?
b)What is the probability of selecting a case to analyze and finding the crime was committed by someone less than 40 years old?
c)What is the probability of selecting a case that involved a violent crime or an offender less than 20 years old?
d)Given that the violent crime is seleced for analysis, what is the probability the crime was committed by a person under 20 years old?
e)Two crimes are seleced by a Judge. What is the probability both are violent crimes?

a)C(150,82)
B)C(150,114)
C)C(150,82) * C(150,12)
D)C(150,27)
E)C(150,82) * C(150,81)

Sorry for all the problems. I just want to make sure I am doing these right. Thanks

Looks like it wouldn't let me insert a table. Here is what five should look like:

Under 20 | 20 to 40 | 40 or older | Total
Violent 27 41 14 82
Nonviolent 12 34 22 68
Total 39 75 36 150

For number 2, there are five different possible ways to use the three cars:
(1,3,5);(1,4,4);(2,2,5);(2,3,4);& (2,4,3).
Now you must figure the number of ways to distribute the nine people for each case.

So it would be

C(2,1) * C(4,3) * C(5,5) * C(2,1) * C(4,4) * C(5,4) * C(2,2) * C(4,2) * C(5,5) * C(2,2) * C(4,3) * C(5,4) * C(2,2) * C(4,4) * C(5,3)?

How about the other problems? Are they ok?

Originally Posted by kenshinofkin

So it would be

C(2,1) * C(4,3) * C(5,5) * C(2,1) * C(4,4) * C(5,4) * C(2,2) * C(4,2) * C(5,5) * C(2,2) * C(4,3) * C(5,4) * C(2,2) * C(4,4) * C(5,3)?

NO!
It would be C(9,1)C(8,3)C(5,5)+C(9,1)C(8,4)C(4,4)+C(9,2)C(7,2) C(5,5)+etc...
Note that the last term is each product is 1, so we don't need it.