KeepCalmAdi wrote:
vitaliyGMAT wrote:
Very interesting and, I'd like to say, not an easy question.
Now let's go to our options.
(1) Coach Jackson could choose exactly 20 different teams
That means nCr = 20, where n is our total number of players and r is # of layers we need to choose. The only thing we can actually do is to plug in different values of n and r to get 20 combinations. The most obvious choice is n=20, r=1 and n=20, r = 19. nCr = nCn-r = 20C1=20C19=20. After some time of calculations we’ll find that 6C3 is also equal to 20.
\(6C3 = \frac{6*5*4}{3*2*1} = 20\).
We have three possible values of r = 1, 3, 19.
The phrase "Jackson will choose at least two players" means that 2=< r =< n =< 20. That limits our options to r=3 or r=19. But even two choices means that this option is not sufficient.
(2) At least two players at the tryout will not be chosen.
That means that our r =< n – 2. Said alone without additional data – insufficient.
(1) & (2) taken together: means that our r is in the interval:
2=< r =< 20-2
2 =< r =< 18
And we have only one option left r=3. Sufficient.
Is it really a 600 level question?
Isn't n= 6 for r=3 ?
Can we use n= 20 while solving for 2<=r<=n-2 ?
Hi
I can't get the gist of your question.
20 is not the total number of players we need to choose from (n) but the number of combinations of possible team formations. This can be achied when n=20 and n=6.
In 2<=r<=n-2 our max n is 20 and we are discarding second option r=19, because r<=n<=18.