Hello, Given all the combinations of 6 positive integers from 1 to 35, c(35,6)=1,623,160 such sets.
Question 1: Selecting two random sets from the above. How can we calculate the number of combinations that *both* selected sets intersect the 1623160 combinations in *at-least* 3 elements? Since the two selected sets may intersect from 0 to 5 elements then we can have 6 such cases. I can solve this by using a "brute force" algorithm, but there must be some equation than can do this easily. Examples using the above algorithm: If intersection is 0 points: Answer is: 158,290 sets If intersection is 1 points: Answer is: 155,990 sets If intersection is 2 points: Answer is: 150,710 sets If intersection is 3 points: Answer is: 142,785 sets If intersection is 4 points: Answer is: 131,059 sets If intersection is 5 points: Answer is: 112,105 sets Question 2: The same as above but we select three random sets and we are looking for combinations that belong to *all three* sets. Now this becomes more complicated because the intersections can vary. Does a better way exist to answer this kind of question. Any ideas or directions will be very much appreciated. Thank you, Socratis . . ================================================================= Instructions for joining and leaving this list, remarks about the problem of INAPPROPRIATE MESSAGES, and archives are available at: . http://jse.stat.ncsu.edu/ . =================================================================
