



 
Hi Steve, If I understand the question right, if the first 4 numbers are 1,2,3,4 then none of the other collections of 4 numbers can contain a 1, 2, 3, or 4. The number of ways to choose a collection of k things from a collection of n distinct things is "n choose k", also denoted C(n,k), which turns out to equal n!/(k!(nk)!). Let's solve your problem with 12 numbers. There are 12 choose 4 ways to pick the four numbers for the first combination. This leaves 8 numbers from which to choose 4 for the second combination, and then 4 numbers from which to choose (all) 4 for the third combination. But the answer isn't (12 choose 4) times (8 choose 4) times (4 choose 4) because the order of the combinations is not important. For example, {1,2,3,4}, {5,6,7,8}, {9,10,11,12} is the same collection of combinations as {9,10,11,12}, {1,2,3,4}, {5,6,7,8}. Any collection of 3 combinations can appear in 3! orders, so the quantity (12 choose 4) times (8 choose 4) times (4 choose 4) counts each collection of combinations 3! times. The answer is (12 choose 4) times (8 choose 4) times (4 choose 4) divided by 3!. Victoria  


Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. 