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The 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 13C3-1-8C3
ill work it out
that wasnt right
Hmm okay maybe it was wrong i am not sure....
would it be a permutation
No i think it'll be a combination only.
One task is given to one person only right?
it just says how many ways can the 3 tasks be assigned so that both women and men are given assignments
is that a typo? shouldn't it be 8 men and 5 women
total 13 members and 3 total tasks
Is 220 the answer?
shankvee had the right idea :) but your numbers don't add up ....8+3 = 11 not 13
leave u cant read 8+5
?? 13C3 - 8C3 - 5C3 = 286 - 56 - 10 = 220
sorry thought you were messing with me i am so sorry but 220 is not right
maybe it is permutations then.... 13P3 - 8P3 - 5P3 = 1320
you are a genius
It was permutations
I am so sorry for my comment earlier I had some punk on here earlier and thought you were messing with me
I am very very sorry
no prob, just noticed a typo in your question post
I am so sorry
I have tried the problem 40 times
do you know the formulas for combinations and permutations? most calculators will do them for you as well
somewhat...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
I have a few more permuations and maybe a combo i am going to post that are two part questions....3-5 of these have stumped me, most i can work through
Should 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?