A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
How many 4person committees are possible from a group of 10 people if either Mary OR Jim (but not both) must be on the committee?
anonymous
 4 years ago
How many 4person committees are possible from a group of 10 people if either Mary OR Jim (but not both) must be on the committee?

This Question is Closed

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.2\[{2\choose 1}{8\choose 3}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0a. 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
 4 years ago
Best ResponseYou've already chosen the best response.2I assume Mary and Jim are part of the 10 people.

Zarkon
 4 years ago
Best ResponseYou've already chosen the best response.2if 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
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.2\[{2\choose 1}{8\choose 3}=2\times\frac{8!}{3!(83)!}=112\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks Zarkon  much appreciated. If you have time, would you mind taking a look at some of the other questions I have posted? Thank you!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.