The easiest way to solve this question is to pretend that neither of them exist.

So, we have 5 people and we want to make a group of 4.

5C4 = 5!/4!1! = 5*4!/4! = 5

Choose (a)

Thank you very much for your explanation, stuart... everytime you explain something it seems so easy, simple and straightforward. TYSM for all your help. i got another Q on powerprep which i am posting on new thread. could u pls explain tht as well?
TIA

Stuart Kovinsky wrote:The easiest way to solve this question is to pretend that neither of them exist.

So, we have 5 people and we want to make a group of 4.

5C4 = 5!/4!1! = 5*4!/4! = 5

Choose (a)

I agree with Desi, we love your explanations Stuart!

Stuart,

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!

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!

Stuart Kovinsky wrote:

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!

You're awesome!

:roll:

thanx a million

That was quick. Thanks Stuart .