** smileqx asks**:

There are 50 kids trying out for 7 basketball teams. The organizer decides to randomly pick 7 kids to be team captains and then let them pick thier own teams. How many different ways can the organizer assign the captains?

**Best answer:**

*Answer by DavidK93*

Use the combinatorical operation 50C7, or “from 50, choose 7.” This is equal to 50! / (7! * (50 – 7)!) = 50! / (7! * 43!). The exclamation point is the factorial operation, where n! = n*(n – 1)*(n – 2)*…*(2)*(1). So 50! is the product of the first 50 positive integers. But since you’re dividing by 43!, you eliminate most of those factors and you should be able to complete the computation with your calculator.

This is a problem of pick 7 from 50 where the order of selection is irrelevant. This type of problem is known as a combination and is given by the formula

C(50,7) = 50!/(7! * (50 – 7)!)

where 50! = 50*49*48* . . . *3*2*1

50! /43! = 50*49*48*47*46*45*44

7! = 7*6*5*4*3*2

so the solution is

50*49*48*47*46*45*44/7*6*5*4*3*2

50C7

that is

= 50*49*48*47*46*45*44/7! ways

he just have to select 7 persons out of 50 who will be captains

50!/43!7!