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Question: What is the simple algebraic formula for working out the number of combinations based on the following criteria:
combinations = n choose r (e.g. 50 balls and you choose 5 balls)
filtered_1_combinations = remove combinations where r has numbers x1, x2, x3, x4, x5, x6 (e.g. 1,5,12,24,44,45)
filtered_2_combinations = remove combinations from filtered_1_combinations where combinations has more than 2 odds numbers
filtered_3_combinations = remove combinations from filtered_2_combinations where combinations has between y1 and y2 consecutive numbers (e.g. if y1 = 1 and y2 = 3 then an accepted combination is 1,2,3,5,7 but not 1,2,3,4,7)
Hi this is n choose r . And defined as factorial(n)/(factorial(n-r)* factorial(r)) = n!/((n-r)! r!)
For example 5c2 = 5! / ( (5-2)! * 2!) = (5*4*3*2*1)/(3*2*1 * 2*1) = 10
Hi there
C (n,r)= n!/((n-r)!×r!) COMBINATION FORMULA
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(50/5)=50!/(50-5)!*5!
=(50*49*48*47*46*45!)/45!*5*4*3*2*1
=20*49*47*46
=2118760