Combinations question i think?

Question

Two balls are to be selected without replacement from a bag containing one red, one blue, one green, one yellow, and one black ball. How many points are there in the sample space?

Answer 1

#combinations of n items m at a time = n!/(n-m)!

For this problem n = 5 and m = 2

therefore: 5*4*3*2*1/3*2*1

Think of it this way.

When you pick the first ball, you can pick one of five possible balls. After you pick the first ball, there are only four balls left (i.e., no replacement). Therefore, you only have four choices for the second ball. The total number of combinations (i.e., sample space) is 5 * 4 = 20.

If this doesn't make sense, then make a table like this:

1st ball  = red

2nd ball = blue, green, yellow, or black

This gives 4 combinations of two balls or 4 members of the sample space. Now change the color of the first ball.

1st ball = blue

2nd ball = red, green, yellow, or black

This gives 4 more combinations. The sample space now has 8 members. Continue through all 5 choices of the color for the first ball and you will arrive at a sample space of 5 * 4 = 20 combinations.

Note 1: If you have replacement, then you put the first ball back in the bag. Therefore, you have 5 choices for the first ball, and also 5 choices for the second ball. This gives a sample space of 5 * 5 = 25.

Note 2: These calculations assume that choosing a red ball first and a blue ball second is a different result as choosing a  blue ball first and red ball second. This is described as 'order matters'. The answer is different (i.e., fewer possible combinations) if order doesn't matter. However, most problems like this want you to assume that order counts.

Note 3: If order doesn't matter then the formula is:

# combinations  =  n!/(m! * (n-m)!)

# combinations = 5!/(2! * 3!) = 10.