anonymous
  • 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?
Mathematics
jamiebookeater
  • jamiebookeater
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Zarkon
  • Zarkon
\[{2\choose 1}{8\choose 3}\]
anonymous
  • 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
  • Zarkon
I assume Mary and Jim are part of the 10 people.

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anonymous
  • anonymous
I would assume so.
Zarkon
  • 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
  • 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
  • Zarkon
\[{2\choose 1}{8\choose 3}=2\times\frac{8!}{3!(8-3)!}=112\]
anonymous
  • 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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