Two members of a club are to be selected to represent the club at a national meeting. If there are 190 different possible selections of the 2 members, how many members does the club have?
20
Problem Solving
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Answer is 20
First member can be selected in N ways and the second member can be selected in (N-1) ways , but you can arrange the two in 2! ways
190= N*(N-1)/2! ................. N(N-1) = 380 , N = 20
First member can be selected in N ways and the second member can be selected in (N-1) ways , but you can arrange the two in 2! ways
190= N*(N-1)/2! ................. N(N-1) = 380 , N = 20
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Here is how we arrive at n(n-1).
General Form
n! = n*(n-1)*(n-2)*(n-3)*(n-4)*(n-5)*...*1
I can shorten the above representation as follows:
n! = n*(n-1)*(n-2)! (by adding the "!" factorial notation)
C(n,r) = n!/[r!*(n-r)!]
C stands for combination
In our case, the problem is something like this
C(n,2) = n!/[2!*(n-2)!] ... I
Expanding the numerator in I and taking hint from the general form we get
= [n*(n-1)*(n-2)!] / [2! * (n-2)!]
Now we have (n-2)! in both the numerator and denominator which we can get rid off. Doing this we would have
= [n*(n-1)]/2! ... II
But from the problem II = 190 (R.H.S)
Therefore,
=> [n*(n-1)]/2! = 190
=> [n*(n-1)] = 380
=> n*(n-1) = 380
I am sure you are in control from here on.
Best of luck.
General Form
n! = n*(n-1)*(n-2)*(n-3)*(n-4)*(n-5)*...*1
I can shorten the above representation as follows:
n! = n*(n-1)*(n-2)! (by adding the "!" factorial notation)
C(n,r) = n!/[r!*(n-r)!]
C stands for combination
In our case, the problem is something like this
C(n,2) = n!/[2!*(n-2)!] ... I
Expanding the numerator in I and taking hint from the general form we get
= [n*(n-1)*(n-2)!] / [2! * (n-2)!]
Now we have (n-2)! in both the numerator and denominator which we can get rid off. Doing this we would have
= [n*(n-1)]/2! ... II
But from the problem II = 190 (R.H.S)
Therefore,
=> [n*(n-1)]/2! = 190
=> [n*(n-1)] = 380
=> n*(n-1) = 380
I am sure you are in control from here on.
Best of luck.