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Mathematics 7 Online
OpenStudy (anonymous):

how many 6 person committees are possible from a group of 8 people a)there are no restrictions b)both jim and mary must be in the group c)either jim or mary (not both) must be on the committee

OpenStudy (anonymous):

a) 8C6

OpenStudy (anonymous):

if both jim and mary are in it, then you have 4 open places out of 6 so second one is \[\dbinom{6}{4}=\frac{6\times 5}{2}=15\]

OpenStudy (anonymous):

both of the responese are great. Thank you ShaileshAR and Satellite

OpenStudy (anonymous):

so how would I compute C??

OpenStudy (anonymous):

if jim must be in it and mary must not be, then you have 5 spots to fill and 6 people to choose from

OpenStudy (anonymous):

so that one is \(\dbinom{6}{5}=6\) the other is identical, so your answer should be \(2\times 6\)

OpenStudy (anonymous):

Ok i get it. Thanks for breaking it down exactly

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