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- anonymous

How many 4-person committees are possible from a group of 10 people if either Mary OR Jim (but not both) must be on the committee?

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- anonymous

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- Zarkon

\[{2\choose 1}{8\choose 3}\]

- anonymous

a. there are no restrictions
no. of ways = 12C4 = 495
b. both jim and mary must be on the committee
no.of ways = 10C2 = 45
c. either jim or mary (not both) must be on the committee
no. of ways when jim on the committee but mary not = 10C3 = 120
no. of ways when mary on the committe but jim not = 10C3 = 120
total ways = 120+120 = 240

- Zarkon

I assume Mary and Jim are part of the 10 people.

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- anonymous

I would assume so.

- Zarkon

if there are 10 total including mary and jim them my answer is correct
if it is 10 people plus mary and jim (total of 12 people) then Icewinifredd's answer is correct

- anonymous

I did the first two parts of the question myself and got them correct but not with those answers.
No restrictions = 210 Formula: |dw:1326951001243:dw|
Both Jim and Mary = 28 Formula: |dw:1326951078166:dw|
Those answers are correct as marked by MyMathLab. I believe the last formula should look something along the lines of: |dw:1326951155771:dw| but I am not sure at all what goes in the bottom to make it work. I only have one shot left at this answer, so I just wanted to be sure that it is right. Thanks guys.

- Zarkon

\[{2\choose 1}{8\choose 3}=2\times\frac{8!}{3!(8-3)!}=112\]

- anonymous

Thanks Zarkon - much appreciated. If you have time, would you mind taking a look at some of the other questions I have posted? Thank you!

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