How to to count the sample space if sampling is done without replacement.?

Question

The text book I have doesn't clearly explain how the formula can be calculated. Example, if there is a population of 6 and a sample of 3, the answer is 20. Can someone please help explain to me how 20 was derived? Thanks.

Answer 1

The sample space is the set of all possible outcomes. In this case, there are 20 possible ways to sample 3 out of 6 without replacement. The mathematical operation is "choosing," or nCr, which means "from n, choose r." nCr = n! / (r!*(n - r)!), where ! is the factorial operation. In this case, we have 6C3 = 6! / (3!*(6 - 3)!) = 6! / (3!*3!) = 720 / (6*6) = 720 / 36 = 20.

This is not the sample space itself, but its size. If the six people were called Adam, Betty, Carol, David, Ernie, and Fran, then the sample space would actually be something like {(Adam, Betty, Carol), (Adam, Betty, David), (Adam, Betty, Ernie), ... , (David, Ernie, Fran)}.

Answer 2

i'm no longer likely particular what the questions are soliciting for yet.... there is 3 slots for any quantity Then 2 slots for the two of the two different numbers Then 3 slots for any letter. 3 * 2 * 3 = 18 combinations of license plates. i can't rather make a tree diagram at here... The danger of having a registration quantity plate with R could in basic terms be a million/3, for the reason that there are 3 letters to compliment from. ----------------------------------- there's a 4/39 risk of choosing a defective pair 4/39 For the 1st pair. Now there are purely 38 denims to compliment from, and there could be purely 3 defective pairs left, for the reason that we are calculating the prospect of having the two defective. 4/39 * 3/38 = 12/1482 = 2/247