# Thread: What are the chances?

1. ## 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.

2. 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: .$\displaystyle {14\choose9} \:=\:2002$ possible outcomes.

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

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

3. Originally Posted by Soroban
Hello, BG5965!

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

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

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

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

4. Hello, BG5965!

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

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

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