preciousrain7 wrote:
I have a question:
why isnt the answer 5*4*3*2, since you have 5 people to choose from for the first individual, then 4 people to choose from for the second individual, 3 people to choose from for the 3rd individual and 2 people to choose from for the 4th individual. Though I realize the answer is not one of the choices. THANKS!
Your solution gives us the number of
ordered groups of 4 we can make out of the 5 people. That's a permutations issue, not a combinations one. Your method will count the same group of people multiple times (e.g. if we call the people A, B, C, D and E, you've counted ABCD, ABDC, ACDB, ACBD, ... as different groups).
We use the combinations formula when we're counting unordered subgroups. We use the permutations formula when we're counting ordered subgroups.
For example:
5 people are trying out for a band, 4 of whom will be selected. How many different groups of 4 can be chosen.
We don't care about the order in which we choose them, so we use the combinations formula:
5C4 = 5!/4!1! = 5*4*3*2/4*3*2 = 5
(nCk = n!/k!(n-k)! is the basic combination formula)
or
5 people are trying out for a marching band, 4 of whom will be selected. In how many different orders can the 4 chosen members march?
Here we do care about the order in which we put the entities, so we use the permutations formula:
5P4 = 5!/1! = 5*4*3*2 = 120
(nPk = n!/(n-k)! is the basic permutation formula)
Hope that helps!
