My Math Forum Determining the number of combinations for n paired events

 July 24th, 2009, 11:17 AM #1 Newbie   Joined: Jul 2009 Posts: 1 Thanks: 0 Determining the number of combinations for n paired events Hi All, I'm trying to come up with a formula that will calculate the possible combinations of a set of paired events. To give an example: There are 6 events, A1, A2, B1, B2, C1, C2 A2 can not occur unless A1 has occured, but A2 does not necessarily need to occur immediately after A1, similarly for B1/B2 and C1/C2 A1 B1 and C1 can occur in any order All 6 events must occur So one possible combination aould be B1, C1, C2, A1, B2, A2 How many such combinations are there? I can work it out by hand but can't find the formula that would allow me to calculate it or even extend this to more than 3 paired events or figure out the combinations for n grouped events (so n groups of x events for a total of n*x events...in the initial case n=3 x=2) Can anyone can point me in the right direction to get me started on solving this? Thanks, Nick
 August 2nd, 2009, 04:04 AM #2 Member   Joined: Jul 2009 Posts: 34 Thanks: 0 Re: Determining the number of combinations for n paired events Let's first look at the A-events. If you chose two of the six places then you can put the A-events on that places in exactly 1 way. So there are ${6 \choose 2}$ ways to place the A-events. For the B-events then there are still ${4 \choose 2}$ ways and for the C-events ${2 \choose 2}$ ways. The total is then ${6 \choose 2}{4 \choose 2}{2 \choose 2}$ combinations. The generalisation is: $\prod_{i=0}^{n-1} {x(n-i) \choose x}$

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