## tim93 3 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?

1. tim93

a) = 22 choose 5 ???

2. satellite73

first one is straightforward yes? no restrictions it is \[\dbinom{22}{5}\]

3. satellite73

yup

4. cinar

|dw:1341688571130:dw|

5. tim93

b) 10 choose 2 plus 12 choose 3

6. satellite73

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

7. satellite73

counting princple times not plus

8. tim93

sorry, times for all combinations, not plus

9. cinar

|dw:1341688655070:dw|

10. satellite73

last one will require an addition because it is an "or" statement no clowns or one clown or two clowns

11. tim93

aha, i was thinking along those lines...thanks guys, see you next chapter

12. satellite73

what @cinar said

13. rachelk09

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?