Combination problem: coach has 5 good hitters and 4 poor

[tdjb]

My guess is I'm just over thinking this but I'm looking for some help with a homework problem. The problem from the text is:

"If a baseball coach has 5 good hitters and 4 poor hitters on the bench and chooses 3 players at random, in how many ways can he choose at least 2 good hitters?"

I am familiar with the formula n!/(n-r)!r!, but my problem is actually setting the initial problem up. I'm pretty sure that since the problem states "at least 2" I'm going to need to add 5C2 and 5C3 because that will give me the sum of those combination, but how do I incorporate the other "4 poor hitters" into it?
Is it as simple as just doing (5C2)(4C1) + (5C3)(4C0) ?

Thanks in advance

 

[Denis]

Re: Combination problem...

Easier to look for the bad guys:

1 2 3 4 5 6 7 8 9 : 1 2 3 4 5 = good
Bad picks(3 bads or 2 bads); example:
7 8 9, 1 7 8

 

[soroban]

Re: Combination problem...

Hello, tdjb!

A baseball coach has 5 good hitters and 4 poor hitters and chooses 3 players at random.
In how many ways can he choose at least 2 good hitters?

Is it as simple as just doing: (5C2)(4C1) + (5C3)(4C0) ? . . . . . Yes!

There are two cases to consider . . .

2 Good and 1 Poor
\(\displaystyle \text{There are: }\:\left(_5C_2\right)\left(_4C_1) \;=\;10\cdot4\;=\;40\text{ ways.}\)

3 Good and 0 Poor
\(\displaystyle \text{There are: }\:\left(_5C_3\right)\left(_4C_0) \;=\;10\cdot1 \;=\;10\text{ ways.}\)

\(\displaystyle \text{Therefore, there are: }\;40 + 10 \:=\:50\text{ ways.}\)

 

[tdjb]

Re: Combination problem...

Thank you both for the help, I appreciate it. My guess is I'll probably be posting up some more questions soon. For whatever reason I'm having a hard time grasping this stuff :? .