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Math Help - What are the chances?

  1. #1
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    What are the chances?

    A school principal needs to employ 9 staff members. There are 4 lab assistants and 10 science teachers. How many ways can he select 9 staff members if he needs at least one lab assistant?

    I think I need to use combination's or permutations or something like that. My first attempt got me an answer above 1000, so I'm not sure if I'm right. Please help.
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  2. #2
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    Hello, BG5965!

    A school principal needs to employ 9 staff members.
    There are 4 lab assistants and 10 science teachers.
    How many ways can he select 9 staff members if he needs at least one lab assistant?
    There are: . {14\choose9} \:=\:2002 possible outcomes.


    The opposite of "at least one LA" is "no LA's" . . . that is, "all ST's."
    . . There are: . {10\choose9} = 10 ways to choose all ST's.


    Therefore, there are: . 2,002 - 10 \:=\:1,992 ways to have at least one LA.

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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, BG5965!

    There are: . {14\choose9} \:=\:2002 possible outcomes.


    The opposite of "at least one LA" is "no LA's" . . . that is, "all ST's."
    . . There are: . {10\choose9} = 10 ways to choose all ST's.


    Therefore, there are: . 2,002 - 10 \:=\:1,992 ways to have at least one LA.

    Thanks, except I don't know what the number in the brackets {14\choose9} and {10\choose9} mean. Can you explain further? Thankyou
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  4. #4
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    Hello, BG5965!

    Thanks, except I don't know what the number in the brackets {14\choose9} and {10\choose9} mean.
    Can you explain further?
    Those are "combinations" . . . are you familiar with them?

    They are also written: . _{14}C_9\,\text{ and }\,_{10}C_9

    The first one is: . {14\choose9} \:=\:\frac{14!}{(9!)(5!)} \:=\:\frac{14\cdot13\cdot12\cdot11\cdot10}{5\cdot4  \cdot3\cdot2\cdot1} \:=\:2,002

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