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A committee of size five is selected from a group of ten clowns and twelve lion tamers. (a) How many di erent committees are possible? (b) How many committees are possible if there must be exactly two clowns on the committee? (c) How many committees are possible if lion tamers must outnumber clowns on the committee?

Mathematics
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a) = 22 choose 5 ???
first one is straightforward yes? no restrictions it is \[\dbinom{22}{5}\]
yup

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Other answers:

|dw:1341688571130:dw|
b) 10 choose 2 plus 12 choose 3
which really just restates the question because \(\binom{22}{5}\) reads 22 choose 5" and means the number of ways to choose 5 from 22
counting princple times not plus
sorry, times for all combinations, not plus
|dw:1341688655070:dw|
last one will require an addition because it is an "or" statement no clowns or one clown or two clowns
aha, i was thinking along those lines...thanks guys, see you next chapter
what @cinar said
Is there any possible way you can show how to do out the (22 5) I'm a little confused about that part of things?

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