SOLUTION: Can you please tell me if I have this right? If not, can you please explain... Thanks sooo much! Assume a class has 30 members. How many committees of three people can be chosen

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Question 245966: Can you please tell me if I have this right? If not, can you please explain... Thanks sooo much!
Assume a class has 30 members. How many committees of three people can be chosen?
I have:
30*29/3=290
Found 2 solutions by stanbon, Alan3354:
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
Assume a class has 30 members. How many committees of three people can be chosen?
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A committee is a group of people so this is a combination problem.
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30C3 = 30!/[(30-3)!*3!]
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= (30*29*28)/(1*2*3)
= 4060 committees
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Cheers,
Stan H.

Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website!
Assume a class has 30 members. How many committees of three people can be chosen?
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30/3 = 10 committees.
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If you meant in how many different ways can a committee of 3 be chosen:
The 1 choice is 1 of 30.
Then 1 of 29, then 28.
--> 30*29*28 = 24360.
But, a committee of A, B and C is the same as C, A and B, so it's necessary to divide by 6 (3*2*1 = 6)
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24360/6 = 4060 different combinations.