# Counting principle

• May 4th 2010, 07:14 PM
ihavvaquestion
Counting principle
Eight girls have decided to pair to partake in the Physics Lab. The work space is big enough so that everyone will be working at once. How many different ways can the lab pairs be formed?
• May 4th 2010, 09:29 PM
Anonymous1
Quote:

Originally Posted by ihavvaquestion
Eight girls have decided to pair to partake in the Physics Lab. The work space is big enough so that everyone will be working at once. How many different ways can the lab pairs be formed?

${8\choose 2} = \frac{8!}{(8-2)!2!} = 28$
• May 4th 2010, 09:35 PM
ihavvaquestion
That is how I originally answered the question. Here is what my prof responded with:

If you did simply 8C2, you would be calculating how many choices you would have for using just one pair of workers while all the rest of the students sat out. The problem asks that you take all 8 of the students, break them into pairs of two, and all four pairs would be working on the experiement at the same time.

He also wrote:

No. One thing you might do is to start with just four people, A, B, C, and D. Write out eactly how many ways you could form them all into teams of two. Example AB, CD would be one way. List the rest of them and see if the counting method that you are trying to use actually produces this number.

Then I thought that maybe....
There are 7 possible partners for the first pair, then 5 possible partners for the next pair, 3 for the third pair, then one left for the last pair. Would that mean there are 7*5*3*1=105 different groups?

Any insight would be great, cause i'm stumped.

• May 4th 2010, 09:53 PM
Anonymous1
Quote:

Originally Posted by ihavvaquestion
That is how I originally answered the question. Here is what my prof responded with:

If you did simply 8C2, you would be calculating how many choices you would have for using just one pair of workers while all the rest of the students sat out. The problem asks that you take all 8 of the students, break them into pairs of two, and all four pairs would be working on the experiement at the same time.

He also wrote:

No. One thing you might do is to start with just four people, A, B, C, and D. Write out eactly how many ways you could form them all into teams of two. Example AB, CD would be one way. List the rest of them and see if the counting method that you are trying to use actually produces this number.

Then I thought that maybe....
There are 7 possible partners for the first pair, then 5 possible partners for the next pair, 3 for the third pair, then one left for the last pair. Would that mean there are 7*5*3*1=105 different groups?

Any insight would be great, cause i'm stumped.

Your right on both accounts. I wasn't splitting all of them into groups.

In this case the answer would be $7\cdot 5\cdot 3\cdot 1.$