Problem Solving

2 membersof a club are to be selected to represent the club meeting. If theres are 190 different possible solutions of the two members. How many members does the club have?

it is acombination prob
nC2= fact (n)/fact(2)*fact(n-2)=190

ie fact n / fact n-2 =380

ie n(n-1) = 380 = 20*19

ie n=20

2 membersof a club are to be selected to represent the club meeting. If theres are 190 different possible solutions of the two members. How many members does the club have?

This is the case of reverse construction

xC2 = 190
x! /2! *(x-2)! = 190
Or x! / (x-2)! = 380
x * x-1 = 380
20 * 19 = 380
Hence x = 20

I get it up to the part when, x!/x-2=x*x-1, why was it changed to x*x-1 so if it was x!/x-3 would it still be x*x-1?? please explain this . SO that I can know how to do the reverse contruction on similar problems in the future.

I would appreciate it

I get it up to the part when, x!/x-2=x*x-1, why was it changed to x*x-1 so if it was x!/x-3 would it still be x*x-1?? please explain this . SO that I can know how to do the reverse contruction on similar problems in the future.

Sample this:

x factorial or "x!" can also be written as x*(x-1)! or 5! can also be written as 5*4!
Here is the reason why:
5! = 5.4.3.2.1
and 5*4! = 5* (4.3.2.1) = 5.4.3.2.1

Thus when x!/(x-2)! (substitute 5 for x)
5! / (5-2)! = 5*4!/3!
=5*4*3!/3!
=5*4

Hope this is helpful

very very very very helpful. Thank you