2 members
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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
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
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- Joined: Tue Mar 27, 2007 1:29 am
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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
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