Combination Formula with a lockout twist

  • Thread starter Calcunaut
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  • #1
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Combination Formula with a "lockout" twist

Hi! I am trying to figure out all possible combonations for 6 items among a group of 18 choices. So I turn to my old friend C(n,r) to calculate where n=18 and r=6. "But WAIT!" I tell you before you hastily begin scribbling, "There is a twist..." You see my problem is that the items are divided up into 6 groups, with 3 choices in each group. Once a choice has been made in a group for the combination the other 2 in the group are unavailable, or "locked out" of the rest of the combination. The order doesn't necessarily matter but a choice must be selected from each of the six groups. Here's a visual representation:

A B C
1 A1 B1 C1
2 A2 B2 C2
3 A3 B3 C3
4 A4 B4 C4
5 A5 B5 C5
6 A6 B6 C6

If "B1" is selected in a single combination then "A1" and "C1" cannot be apart of the same combination. What is the formula for this and how many possible combinations are there?
 

Answers and Replies

  • #2
Stephen Tashi
Science Advisor
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Do you have 18 distinct items? The answer is [itex] 3^6 [/itex].

Or are you counting {A1,A2,A3,B4,B5,A6} to be the same as {A1,B2,B3,A4,A5,A6}?
 
  • #3
chiro
Science Advisor
4,790
132


If its like the bike locks that are popular, each "ring" is independent and like Stephen said, there will be 3x3x3x3x3x3 combinations (or 3^6).
 

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