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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?
 one year ago
 one year 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?
 one year ago
 one year ago

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satellite73Best ResponseYou've already chosen the best response.0
first one is straightforward yes? no restrictions it is \[\dbinom{22}{5}\]
 one year ago

tim93Best ResponseYou've already chosen the best response.0
b) 10 choose 2 plus 12 choose 3
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
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
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
counting princple times not plus
 one year ago

tim93Best ResponseYou've already chosen the best response.0
sorry, times for all combinations, not plus
 one year ago

satellite73Best ResponseYou've already chosen the best response.0
last one will require an addition because it is an "or" statement no clowns or one clown or two clowns
 one year ago

tim93Best ResponseYou've already chosen the best response.0
aha, i was thinking along those lines...thanks guys, see you next chapter
 one year ago

rachelk09Best ResponseYou've already chosen the best response.0
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?
 one year ago
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