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June 21st, 2017, 04:20 PM  #1 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 561 Thanks: 79  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 ?

June 21st, 2017, 07:02 PM  #2 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,602 Thanks: 816 
yes maybe written a bit sexier as $\prod \limits_{k=1}^{19} \dbinom{(20k)10}{10}$ 
June 22nd, 2017, 04:54 AM  #3 
Senior Member Joined: Feb 2010 Posts: 632 Thanks: 103 
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. 
June 22nd, 2017, 05:34 AM  #4  
Senior Member Joined: Dec 2012 From: Hong Kong Posts: 853 Thanks: 311 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics  Quote:
 
June 26th, 2017, 06:33 PM  #5 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 561 Thanks: 79 
The orders of groups matter.


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190, balls, combinations, drawn, groups 
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