Question

how many 3 person committees are possible from a group of 10 people if either jim or mary but not both must be on the committee

Answer 1

Okay, I've figured it out

Answer 2

1. Create two separate groups. Put Jim, Mary in the first group. Put the remaining eight people in the other group:

Group one: Jim, Mary
Group two: The other eight committee members

2. Choose one person from the first group, then choose 2 people from the second group.
2C1 * 8C2

Answer 3

so the answer is 16?

Answer 4

It's not 16. I don't know how you got that bro. Do you not know how to calculate combinations?

Answer 5

Use the combinations formula and multiply both combinations together.

Answer 6

I practically gave you the answer and you still don't know how to calculate it.

Answer 7

Did you figure it out yet bro?

Answer 8

You use the combinations formula

\[nCr = \frac{n!}{r!(n-r)!}\]

Answer 9

i got 36

Answer 10

It's not 36

Answer 11

How are you calculating these bro?

Answer 12

8!/2!6! 8*9/2*1=36

Answer 13

You only calculated one of them

Answer 14

And I don't believe it's right

Answer 15

did u get 28

Answer 16

8C2 = 28

Yes, but what about 2C1

Answer 17

That's also a part of the calculation.

Answer 18

You're supposed to calculate 2C1 x 8C2

Answer 19

64?

Answer 20

That's still not correct.

Answer 21

Are you using the formula I gave you?

Answer 22

Communicate with me faster.

Answer 23

ya i used it for 28 but how do i get the other number

Answer 24

You use the same formula

Answer 25

okay so it would be 28 times 2

Answer 26

Yes, which equals what?

Answer 27

56

Answer 28

56 is the correct answer....finally.

Answer 29

thank you so muich for your time and patience i appreciate it!