Two members of a club are to be selected to....

[fahadmuhammad86]

I know its pretty straightforward question, but for some reason i cant get the right answer. Can any one please solve this question.

Q : 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?

Answer is 20.

[vili_exisu]

Let's assume that there are n members, selecting 2 persons out of n members = [n x (n-1)]/2. This is equal to 190.

You have quadratic expresion that you can solve for n.

[fahadmuhammad86]

Im still a bit confused about the equation you came up with. Can you please explain it as to how you came up with that equation?

[diaconu.oana]

I guess it is easier to think that 190=20*19/2! than solve that quadratic. Think of two slots in which you need to put 2 people. Only that the order of selection does not matter (that is why you divide by 2!). Am I corect?

[aravindc78]

Im still a bit confused about the equation you came up with. Can you please explain it as to how you came up with that equation?

Let number of members in the club be x. So, basically the information tha we have is xC2 = 380.

So if you simplify this, you will get a quadratic equation.

x fac/(2 factorial)(x-2 factorial) = 190
x factorial/(x-2 factorial) = 380

This simplifies to, x(x-1) = 380

x sq - x - 380 = 0. This implies (x-20)(x+19) = 0. Therefore, x = 20.

[RonPurewal]

I know its pretty straightforward question, but for some reason i cant get the right answer. Can any one please solve this question.

Q : 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?

Answer is 20.

if you're not crackerjack at combinatorial formulas, then you should just use the slot method (further details in our strategy guides):

* MAKE THE SLOTS
you need TWO slots, since you are making two selections.

* FILL THE SLOTS IN
if there are N people, then you have N choices for the first person (since you can pick anyone). for the second person, you have (N - 1) choices, since you can't pick the same person you chose the first time.

* DIVIDE BY THE FACTORIAL IF ORDER DOESN'T MATTER
since order doesn't matter here (you're just picking "2 members" - no distinction is made between them), you need to divide by 2!.

therefore, (N)(N - 1)/2! = 190. solving gives N = 20 or -19, of which only the former makes any sense.

--

by the way, it's much easier to PLUG IN THE ANSWER CHOICES on this problem.
just try the numbers in the problem ("work backwards"), and see whether you get 190.

let's pretend the choices were, say, 12, 15, 18, 20, and 25 (since the original poster didn't give the choices).

first try 18, since it's the middle choice.
if N = 18, then, by the same reasoning as above, there are 18 x 17 / 2! = 153 possibilities (note that you don't have to solve any annoying quadratics if you plug in the choices and work backwards).
this number is too small. therefore, 12, 15, and 18 are out. (this is why we do the middle choice first - we can kill 3 answers.)

since 153 isn't that far off 180, try the closer answer, N = 20. that gives 20 x 19 / 2!, or 190. jackpot.