Calculate combinations given certain criteria

[james1000]

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

 

[Deleted member 4993]

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

Please post your problem EXACTLY as it was given to you - verbatim.

Right now - as posted - I cannot figure out your problem.

 

[Romsek]

I think you could benefit from a quick read on generating functions as applied to combinatorial problems.

Maybe this will help. http://math.mit.edu/research/highschool/primes/materials/2018/conf/15-1 Manne.pdf

 

[pka]

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)!)

James, frankly I do not understand any of what you wrote beyond the quoted above.
I whole-heartily Ecco what Mr. Khan asked you to do. Please state some problems that interest you. Be complete in the statements.
Here is a start. The symbol \(\displaystyle \dbinom{N}{k}\) is called the binomial coefficient.
But more commonly that is the number of combinations of \(\displaystyle N\) distinct items choosing \(\displaystyle k\) at a time.
That is evaluated as \(\displaystyle \dbinom{N}{k}=\dfrac{N!}{(k!)((N-k)!)}\)
From what you posted \(\displaystyle \dbinom{50}{5}=\dfrac{50!}{5!\cdot 45!}\)
Here is a great website: LOOK AT THIS You can use that tool to check your calculations and save you time.

 

[JeffM]

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

I too am not sure what you are asking. In the first, I suspect that you are asking how many ways can you select five distinct numbers out of fifty distinct numbers such that at most three of the five selected are odd? Is that correct?

If so, it is numerically unanswerable. For example, if all fifty numbers are odd, there are zero ways to select five numbers at least two of which are even. Unless you specify how many of the fifty are odd, there is insufficient information to compute a numeric answer.

As for what you are trying to ask in the second question, I have no clue whatsoever.