# Arrangements.

• Feb 10th 2010, 05:01 PM
Hapa
Arrangements.
Thanks in advance for the assistance!

1) Farmer John has 9 prize winning cows. How many ways can he choose 3 of his cows to show at the state fair?

Is it as simple as "9 C 3", 9 choose 3, on the caculator?

2) If 3 people need to serve as chaperones on a shool trip, how many ways can they be chosen from the parents of 20 students? (Assume 2 parents per student)

Is this as simple as "40 C 3"?
• Feb 10th 2010, 11:39 PM
Roam
Quote:

Originally Posted by Hapa
Thanks in advance for the assistance!

1) Farmer John has 9 prize winning cows. How many ways can he choose 3 of his cows to show at the state fair?

Is it as simple as "9 C 3", 9 choose 3, on the caculator?

2) If 3 people need to serve as chaperones on a shool trip, how many ways can they be chosen from the parents of 20 students? (Assume 2 parents per student)

Is this as simple as "40 C 3"?

Yes. I think you have to use the $^nC_r$ on your calculator or do it by hand

$^9C_3 = \frac{9!}{3! \times (9-3)!}$

$=\frac{9!}{3! \times 6!} = 84$

In a problem like this there are always two ways of looking at the same selection. One is the number of ways of choosing the objects to be taken, the other is choosing the ones to be left behind. If 3 are chosen then remaining "9-3" are automatically not chosen. Both have to be equal.

$^nC_r= \frac{n!}{n! \times (n-r)!}= \frac{n!}{(n-r)! \times r!}$

$
^nC_{n-r} = \frac{n!}{(n-r)! \times (n-[n-r])!} = \frac{n!}{(n-r)! \times r!}$

Same with the second question.