Number combinations given certain criteria

[Guest]

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. Sorry for lack of algebra as mine is appalling. Appreciate any expert's help!

James

[HallsofIvy]

You are choosing (without replacement?) 5 cards out of 50 (distinct? are they 1 to 50?). Assuming my assumptions are correct, there are 25 even and 25 odd cards. The probability the first is even is 25/50. Then there are 24 even and 25 odd cards. The probability the second card is even is 24/49. In the same way, the probability the third, fourth, and fifth cards are odd are 25/48, 24/47, and 23/46. The probability or "two even, three odd", in that order, is [tex]\frac{25}{50}\frac{24}{49}\frac{25}{48}\frac{24}{47}\frac{23}{46}[/tex][tex]= \frac{[(25)(24)][(25))(24)(23)]}{(50)(49)(48)(47)(46)}[/tex]= [tex]\frac{25!}{23!}\frac{25!}{22!}\frac{45!}{50!}[/tex]. It is not hard to see that "two even three odd" in any order has that probability and that there are [tex]\frac{5!}{3!2!}[/tex] different orders so there probability of "two even and one odd" is [tex]\frac{5!}{3!2!}\frac{25!}{23!}\frac{25!}{22!}\frac{45!}{50!}[/tex].