varun289 wrote:Coach Jackson will choose at least two players for his team from those who try out on Saturday. How many players will Coach Jackson choose?

(1) Coach Jackson could choose exactly 20 different teams.
(2) At least two players at the tryout will not be chosen.

Let us assume there are n players to choose from and coach Jackson will choose r players, where n ≥ r ≥ 2.
We need to determine r.

Statement 1: nCr = 20
Now there is no option but trying out values for n and r to see for what values of nCr = 20
But we can minimize the search time by if we notice that :

• 1. n must be less than or equal to 20.
  2. As the value of nCr is maximum when r = n/2 (n even) or (n + 1)/2 (n odd) and 5C3 = 10, n must be greater than 5.
  3. n cannot be equal to 7, 11, 13, 17, and 19 as that will make nCr a multiple of these prime numbers for any value of r < n.
  4. As soon as you get nCr > 20 for some r, for a particular value of n, stop checking as with increasing value of r (till n/2) nCr will keep on increasing.
  5. We don't need to check for all possible values of r for a particular value of n. Values till n/2 will suffice as after that nCr repeats the same values, i.e 9C1 = 9C8, 9C2 = 9C7 etc

6C1 = 6, 6C2 = 15, 6C3 = 20
8C1 = 8, 8C2 = 28 > 20 ---> No need to check the rest
9C1 = 9, 9C2 = 36 > 20 ---> No need to check the rest
10C1 = 10, 10C2 = 45 > 20 ---> No need to check the rest
...
We can see that our only chance to get a value of 20 only if nC1 = 20, which is possible for n = 20.

So, 20C1 = 20
Hence 20C(20 - 1) = 20C19 will be also equal to 20.

As r ≥ 2, our only options are 6C3 or 20C19
Hence, either r = 3 or r = 19

Not sufficient

Statement 2: This means r < (n - 2)

Not sufficient

1 & 2 Together: Only possible combination is 6C3
Hence, r = 3

Sufficient

The correct answer is C.