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June 21st, 2017, 04:20 PM   #1
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How Many Combinations Are There When 190 Balls Are Drawn Into Groups Of 10?

Balls numbered from 1 to 190 are drawn without replacement into 19 groups of 10. The order within a group doesn't matter. Is the amount of combinations 190 C 10 * 180 C 10 * ... 10 C 10 ?
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June 21st, 2017, 07:02 PM   #2
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yes

maybe written a bit sexier as

$\prod \limits_{k=1}^{19} \dbinom{(20-k)10}{10}$
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June 22nd, 2017, 04:54 AM   #3
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Hmm ... I think you might need to divide this by 19!

Balls numbered from 1 to 4 are drawn without replacement into 2 groups of 2. The order within a group doesn't matter.

I don't believe the answer is 4C2 * 2C2 = 6.

(12)(34), (13)(24), (14)(23) are the only ways. I think you need to divide by 2! since the groups could be rearranged.
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June 22nd, 2017, 05:34 AM   #4
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Quote:
Originally Posted by mrtwhs View Post
Hmm ... I think you might need to divide this by 19!

Balls numbered from 1 to 4 are drawn without replacement into 2 groups of 2. The order within a group doesn't matter.

I don't believe the answer is 4C2 * 2C2 = 6.

(12)(34), (13)(24), (14)(23) are the only ways. I think you need to divide by 2! since the groups could be rearranged.
I think the question simply wasn't posed clearly enough, since it wasn't clear whether the orders of groups count...
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June 26th, 2017, 06:33 PM   #5
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The orders of groups matter.
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