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Mathematics 22 Online
OpenStudy (bloomlocke367):

How would I go about solving this? Marco is putting some sweaters into storage for winter. He has 10 sweaters, and he can fit 6 sweaters into a box. How many different groups of sweaters can Marco pack into a box?

OpenStudy (bloomlocke367):

@rational

OpenStudy (anonymous):

10 C 6?

OpenStudy (bloomlocke367):

?

OpenStudy (bloomlocke367):

@phi

OpenStudy (bloomlocke367):

Can you help? I'm reviewing for my exam, and there's a lot I forgot..

OpenStudy (phi):

I think they are asking how many combinations of 6 sweaters you can pick out of 10 10 Choose 6

OpenStudy (anonymous):

You have 10 sweaters/options and you only can choose 6 How many diff combinations can you have? So the formula for 10 choose 6 is ... \[\frac{n!}{k!(n-k)!}= \frac{10!}{6!4!}\] Where n are the total amt of options and k is the amt you can choose In this case n=10 and k=6

OpenStudy (bloomlocke367):

oh, that makes sense

OpenStudy (bloomlocke367):

210

OpenStudy (anonymous):

yea

OpenStudy (bloomlocke367):

thank you, I totally forgot about that

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