IMO E 10C4 =210

10C4

10!/(10 - 4)! * 4!
10!/6!*4!
10*9*8*7/4*3*2
210

akahuja143 wrote:IMO E 10C4 =210

still confused!

There are four spots on the committee, so for the first spot, any of the ten people could be chosen. There would be nine people left for the second spot, eight for the third, and seven for the fourth. So the first part of the equation is:

10 x 9 x 8 x 7

Because the order in which the people are chosen doesn't matter, this is a combination, not a permutation. So we divide that number (5040) by the number of ways you can arrange a group of 4 people, which is expressed as 4! (4 factorial): 4 x 3 x 2 x 1. So the problem looks like this:

10 x 9 x 8 x 7
4 x 3 x 2 x 1

Since you don't have a calculator, it's easier to reduce before you multiply, so do that first; when you have, the problem will look like this:

5 x 3 x 2 x 7

And that equals 210

grockit_andrea wrote:There are four spots on the committee, so for the first spot, any of the ten people could be chosen. There would be nine people left for the second spot, eight for the third, and seven for the fourth. So the first part of the equation is:

10 x 9 x 8 x 7

Because the order in which the people are chosen doesn't matter, this is a combination, not a permutation. So we divide that number (5040) by the number of ways you can arrange a group of 4 people, which is expressed as 4! (4 factorial): 4 x 3 x 2 x 1. So the problem looks like this:

10 x 9 x 8 x 7
4 x 3 x 2 x 1

Since you don't have a calculator, it's easier to reduce before you multiply, so do that first; when you have, the problem will look like this:

5 x 3 x 2 x 7

And that equals 210

wow, you madethat so easy to understand! thank you so much!