Problem Solving

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?

a. 20
b. 27
c.40
d.57
e. 95





OA A

This can be solved by backsolving.

Lets take C and backsolve:
Number of selections of 2 people from 40 people is
40!/2!38! = 780 - this is too much, we need 190 selections.

Lets take A:
Number of selections of 2 people from 20 people is
20!/2!18! = 190 - correct.

Therefore A is the answer.

baller12q wrote: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?

a. 20
b. 27
c.40
d.57
e. 95





OA A

Hi Ballerq,
Thank you very much for including the answer choices and the OA.
Also thanks for providing interesting questions for all of us to practice and help each other.
However, i would appreciate if u can add the OA within the spoiler.

The way u do it is as follows:
1. click on the Spoiler button. It will add "[" followed by "spoiler]".
2. Now type ur OA text eg. OA: C.
3. Now click on Spoiler button again. it will append "[ " followed by "/spoiler]".

4. Now the OA is not readily visible to any of us. It will be visible only when the mouse pointer is positioned over that area.

Here is an example. Move the mouse pointer over the line below to see the text.

U can see only if u move the mouse pointer over this area

Here is another method.
From the formula for combinations we get
nC2 = n*(n-1)/ 2*1 = 190

so we get a quadratic equation
n^2 -n -380= 0
= n(n -20) +19(n-20) = 0
so n = -19 or +20.

Ppl cannot be -ve.
so n =20.

Nailya wrote:This can be solved by backsolving.

Lets take C and backsolve:
Number of selections of 2 people from 40 people is
40!/2!38! = 780 - this is too much, we need 190 selections.

Lets take A:
Number of selections of 2 people from 20 people is
20!/2!18! = 190 - correct.

Therefore A is the answer.

Backsolving is an awesome approach - but you did too much work!

Using Kaplan's backsolving strategy, we always want to start with either (B) or (D) rather than (C). Let's examine why:

If you start with (C), you have a 1/5 chance of getting the answer on the first shot. As long as you can determine which direction to go (i.e. do you need a bigger answer or a smaller answer), you're then guaranteed to get the correct answer on the 2nd try.

If you start with (B) or (D), you have a 2/5 chance of getting the answer on the first shot. (If you start with (B) and it's too big, you know (A) is correct; if you start with (D) and it's too small, you know (E) is correct). Again, as long as you can determine which direction to go, you're then guaranteed to get the correct answer on the 2nd try.

To summarize:

Start with (C): 1/5 of the time you test 1 choice, 4/5 of the time you test 2 choices.

Start with (B) or (D): 2/5 of the time you test 1 choice, 3/5 of the time you test 2 choices.

On this particular question, if you had started with (B) you would have found that it gave too many combinations and known that (A) is the correct answer without any further work.

Thanks Stuart for the excellent explanation of the strategy.
This should be added to the GMAT resources if it is not already there.

Thank you Stuart!