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January 13th, 2016, 03:19 AM  #1 
Newbie Joined: Feb 2012 Posts: 8 Thanks: 0  Combination containing a pair
I've just been looking and permutations and combinations in a B&C textbook and one of the following exercises stumped me. Q. There's several of each of nine items. How many combinations are there that contain 2 of the same item and two other different items. The permutations for all different items is 9C4 = 9!/(5!4!) = 126 and all 4 the same would be 9. The answer given for the Q is 252 and I can't see how they get this. My solution would be that initially you have 9 choices then the second of the same item is a given and then what's left is the number of combinations of 2 from 7. Soln: 9 * 1 * (7!/(5!/2!)) = 189 (ie. not the answer given 252). What have I missed? 
January 13th, 2016, 05:20 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,527 Thanks: 588 
Wrong forum  try statistics

January 28th, 2016, 11:08 AM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,194 Thanks: 871 
But certainly, just saying "There's several of each of nine items" is not enough. How many of the individual items? That will affect the answer.

January 28th, 2016, 08:02 PM  #4 
Global Moderator Joined: May 2007 Posts: 6,527 Thanks: 588 
I remember seeing the same problem in another forum with the same mistake. To get 252 use $\displaystyle 9\frac{8!}{6!2!}$


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