Hi,
Using my "real life" example below but also considering the numerals used can be anything, how do you calculate the number of combinations given the following criteria:
5 out of 50 numbers chosen (50! /(5!*(50-5)!)
1) less or equal to 2 numbers can be odd (((48/2+1)! /(0!*((48/2+1)-0)!) * ((50/2)! /((5-0)!*((50/2)-(5-0))!)) + (((48/2+1)! /(1!*((48/2+1)-1)!) * ((50/2)! /((5-1)!*((50/2)-(5-1))!)) + (((48/2+1)! /(2!*((48/2+1)-2)!) * ((50/2)! /((5-2)!*((50/2)-(5-2))!))
2) less or equal to 1 number can be from a set of 6 numbers (e.g. 1,4,6,7,22,40) = ((6! /(0!*(6-0)!) * ((50-6)! /((5-0)!*((50-6)-(5-0))!)) + ((6! /(1!*(6-1)!) * ((50-6)! /((5-1)!*((50-6)-(5-1))!))
I can calculate the above for 2) or 3), but not when both meet the criteria together. Appreciate any help!
James
Using my "real life" example below but also considering the numerals used can be anything, how do you calculate the number of combinations given the following criteria:
5 out of 50 numbers chosen (50! /(5!*(50-5)!)
1) less or equal to 2 numbers can be odd (((48/2+1)! /(0!*((48/2+1)-0)!) * ((50/2)! /((5-0)!*((50/2)-(5-0))!)) + (((48/2+1)! /(1!*((48/2+1)-1)!) * ((50/2)! /((5-1)!*((50/2)-(5-1))!)) + (((48/2+1)! /(2!*((48/2+1)-2)!) * ((50/2)! /((5-2)!*((50/2)-(5-2))!))
2) less or equal to 1 number can be from a set of 6 numbers (e.g. 1,4,6,7,22,40) = ((6! /(0!*(6-0)!) * ((50-6)! /((5-0)!*((50-6)-(5-0))!)) + ((6! /(1!*(6-1)!) * ((50-6)! /((5-1)!*((50-6)-(5-1))!))
I can calculate the above for 2) or 3), but not when both meet the criteria together. Appreciate any help!
James