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anonymous
 4 years ago
You have a committee of 13 members, made up of 8 men and 3 women. In how many ways can 3 tasks be assigned so that both men and women are given assignments?
anonymous
 4 years ago
You have a committee of 13 members, made up of 8 men and 3 women. In how many ways can 3 tasks be assigned so that both men and women are given assignments?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The number of ways in which 3 tasks can be distributed such that only men or only women get the job is. Only men 8C3 Only women : 3C3=1 Total ways: 13C3 Hence the answer would be 13C318C3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hmm okay maybe it was wrong i am not sure....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0would it be a permutation

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No i think it'll be a combination only.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0One task is given to one person only right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it just says how many ways can the 3 tasks be assigned so that both women and men are given assignments

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is that a typo? shouldn't it be 8 men and 5 women

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0total 13 members and 3 total tasks

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0shankvee had the right idea :) but your numbers don't add up ....8+3 = 11 not 13

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0leave u cant read 8+5

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0?? 13C3  8C3  5C3 = 286  56  10 = 220

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry thought you were messing with me i am so sorry but 220 is not right

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0maybe it is permutations then.... 13P3  8P3  5P3 = 1320

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I am so sorry for my comment earlier I had some punk on here earlier and thought you were messing with me

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no prob, just noticed a typo in your question post

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have tried the problem 40 times

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0do you know the formulas for combinations and permutations? most calculators will do them for you as well

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0somewhat...i find them confusing and hard to remember also my finite course does not allow calculators during exams so i am trying to do them manually because all of our test questions come from these particular set of problems

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have a few more permuations and maybe a combo i am going to post that are two part questions....35 of these have stumped me, most i can work through

Directrix
 4 years ago
Best ResponseYou've already chosen the best response.0Should this 13 be 11? Or, are the other two committee members neither female nor male? Just wondering. 13 members, made up of 8 men and 3 women. 8+3 =11. I don't think this is a permutation problem. I could be wrong but what about the 13 committee  those two extra members?
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