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tim93
 2 years ago
A committee of size five is selected from a group of ten clowns and twelve lion
tamers.
(a) How many dierent 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?
tim93
 2 years ago
A committee of size five is selected from a group of ten clowns and twelve lion tamers. (a) How many dierent 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?

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satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0first one is straightforward yes? no restrictions it is \[\dbinom{22}{5}\]

tim93
 2 years ago
Best ResponseYou've already chosen the best response.0b) 10 choose 2 plus 12 choose 3

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0which really just restates the question because \(\binom{22}{5}\) reads 22 choose 5" and means the number of ways to choose 5 from 22

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0counting princple times not plus

tim93
 2 years ago
Best ResponseYou've already chosen the best response.0sorry, times for all combinations, not plus

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.0last one will require an addition because it is an "or" statement no clowns or one clown or two clowns

tim93
 2 years ago
Best ResponseYou've already chosen the best response.0aha, i was thinking along those lines...thanks guys, see you next chapter

rachelk09
 2 years ago
Best ResponseYou've already chosen the best response.0Is 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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