Posted: 6/26/2012 2:08:50 PM EDT
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Question.
Say I have 9 numbers (8-6-5-4-20-3-13-9-12). How many different 4 number combinations would I get with those 9 numbers? Oh, and not any one of the 4 numbers in the 4 number combination can repeat while being used in the 4 number combination. So you couldn't have more then one 8, 20 or 12, or any of the other numbers in the 4 number combination at the same time. I just can't figure it out, and google didn't seem to help me. |
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Quoted:
Does the order of the numbers matter or not? i.e, would you consider 6, 5, 4 and 5, 4, 6 to be different? The numbers would be like lottery numbers, so the order wouldn't matter. You just have to have the 4 numbers that were drawn. So I guess I would consider 6,5,4 and 5,4,6 to be the same. |
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Quoted:
Quoted:
Does the order of the numbers matter or not? i.e, would you consider 6, 5, 4 and 5, 4, 6 to be different? The numbers would be like lottery numbers, so the order wouldn't matter. You just have to have the 4 numbers that were drawn. So I guess I would consider 6,5,4 and 5,4,6 to be the same. ok, then this is correct: Quoted:
9 choose 4. n! / (k! (n-k)!) = 9! / (4! (9 - 4)!) = 9! / (4! * 5!) = 362880 / 24 * 120 = 362880 / 2880 = 126 |
