How many different ways can four team members from the group of fifteen be lined up for a photograph?

Question

for six sigma course

Answer 1

I will solve a similar problem using committees of 3 people from a group of 12.

 

There are 12 possibilities for the 1st person, 11 for the second, and then 10 for the third (that is because each time there is one person selected there is 1 less person to choose from.)

 

So the possibilities will be 12 x 11 x 10 = 1320.

 

You aren't finished yet. You need to also consider any possible repeats (i.e. 1,2,3 and 1,3,2 are the same grouping but were selected in different orders.)

 

So of the 3 people choses there are 3! or 3 x 2 x 1 = 6 possibilities. You will need to divide this from the number above to find the number of possibilities, without repeats.