A sports coach intends to choose a team of players from a pool of candidates. The coach wants to be able to have more than 20 but fewer than 25 distinct possibilities for the composition of the chosen team, with at least as many candidates chosen for the team as those not chosen.

Identify the number of candidates in the pool and the number of players on the team that are consistent with the coach’s intentions.
Make only two selections, one in each column.

Number | Candidates in Pool | Players on Team
3
4
5
6
7
8

Solutions:

Hi there, this is an interesting approach.

The question says "with at least as many candidates chosen for the team as those not chosen" and in our case chosen is 5 \geq 2 which are not chosen.
In addition, 7C5 = 7C2 = 7*6/2 = 21 > 20 and 21 < 25.
And for n = 8, there is no k fulfilling the required conditions.
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PhD in Applied Mathematics
Love GMAT Quant questions and running.

nCk = nC(n - k) is the easiest to compute for k = 2 being equal to \frac{n(n-1)}{2}.
You should start with the highest values and go backwards, as 6C3 = 20 and nCk is the highest for k around half of n.
So, you can deduce that n must be greater than 6.
And take advantage of the multiple choice question, as there is just one correct answer.

Otherwise, once you understand Pascal's triangle, it is quite easy to write it down. Having all the values in front of yours eyes, really helps to pinpoint the correct answer.
_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

For choosing teams, you want to keep in mind the combination basics...
you want to choose out of n people, create a team of r. THe number of combinations (NOT permutations) should be between 21-24. That's more than 20 and less than 25.

So, if you try the various combinations the only one that works is out of 7 choose a team of 5.

THat's 7NCR5
7! / (5! 2!) = 6*7/2 = 21

What about 8C4?

8! / (4!*4!) = 5*6*7*8/4 = 30*7*2 = 240

What about 8C5?
8! / (5! 3!) = 6*7*8 / (3*2) = 56

What about 8C6?
8! / (6! 2!) = 7*8 / 2 = 28

What about 8C7?
8! / (7! 1!) = 8

So somehow we don't get in the correct range. You can try for the others but you won't get in range.

6C3 = 6! / (3! 3!) = 4*5*6 / (3*2) = 20
6C4 = 6! / (4! 2!) = 5*6 / 2 = 15

The highest number of combinations will be in the middle. So 6C3 will be the max for 6 people. 4C2 will be the max for 4 people. So if 6C3 only has 20, we know we need to go higher to 7.

7C4 = 7! / (4! 3!) = 5*6*7 / (3*2) = 5*7 = 35 (too high)
7C5 = 7! / (5! 2) = 6*7 / 2 = 21 (just right!)

So if you're familiar with the binomial distribution curve for these combinations and that out of 8 you choose a number in the middle you'll get the highest number of results. Using that you can do an educated guess as to what to try next.
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this is hard. I do not think we need remember the formular.
Just "feel" about correct answer and pick specific number to try. it take me long to do so.