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tim93

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
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  1. tim93
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    a) = 22 choose 5 ???

    • one year ago
  2. satellite73
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    first one is straightforward yes? no restrictions it is \[\dbinom{22}{5}\]

    • one year ago
  3. satellite73
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    yup

    • one year ago
  4. cinar
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    |dw:1341688571130:dw|

    • one year ago
  5. tim93
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    b) 10 choose 2 plus 12 choose 3

    • one year ago
  6. satellite73
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    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
  7. satellite73
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    counting princple times not plus

    • one year ago
  8. tim93
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    sorry, times for all combinations, not plus

    • one year ago
  9. cinar
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    |dw:1341688655070:dw|

    • one year ago
  10. satellite73
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    last one will require an addition because it is an "or" statement no clowns or one clown or two clowns

    • one year ago
  11. tim93
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    aha, i was thinking along those lines...thanks guys, see you next chapter

    • one year ago
  12. satellite73
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    what @cinar said

    • one year ago
  13. rachelk09
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    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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