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Source: — Problem Solving |
Senior | Next Rank: 100 Posts
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by RAGS » Tue Apr 10, 2007 10:50 am
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
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by Cybermusings » Tue Apr 10, 2007 11:18 am
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
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Master | Next Rank: 500 Posts
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Joined: Wed Feb 07, 2007 4:13 pm

ok

by yvonne12 » Wed Apr 11, 2007 4:02 pm
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
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by Cybermusings » Thu Apr 12, 2007 4:36 am
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
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Master | Next Rank: 500 Posts
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thank you

by yvonne12 » Thu Apr 12, 2007 9:08 am
very very very very helpful. Thank you
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