gmatJP wrote:
How many different 5-person teams can be formed from a group of x individuals?
(1) If there had been x + 2 individuals in the group, exactly 126 different 5-person teams could have been formed.
(2) If there had been x + 1 individuals in the group, exactly 56 different 3-person teams could have been formed.
The answer is D and I know how to figure out now but is there any trick to know each question is sufficient without actual compute? cuz its time consuming until I found out e.g. 1) is 9!/5!4!
thanks in advance
If there are x people and we need to choose 5 out of them, we can do that in xC5 ways.
For example, 5C5 = 1
or 6C5 = 6
or 7C5 = 7*6/2 = 21
and so on...
As the value of x increases, the number of ways of selecting will keep increasing since you will have more and more choice to pick from. So for every value of x, you will get a unique number of ways in which you can pick 5 people.
(1) If there had been x + 2 individuals in the group, exactly 126 different 5-person teams could have been formed.
So for some value, there are 126 different ways of picking 5 people. We can find this "some value" which will give us the value of x + 2 and hence of x. This will give us a unique value of xC5.
Hence this statement alone is sufficient.
Note that we don't need to actually calculate the value of x since it will waste our time unnecessarily but if you needed to,
(x+2)C5 = 126 = 2*3*3*7 = 7*6*3 (So x+2 is certainly 7 or higher).
Note that 9C5 will be 9*8*7*6/4*3*2 = 7*6*3
So x+2 = 9 and x = 7
So we need to find 7C5 which is 7*6/2 = 21
(2) If there had been x + 1 individuals in the group, exactly 56 different 3-person teams could have been formed.
Again, x + 1 will have a unique value such that (x+1)C3 = 56.
This means x will have a unique value.
Sufficient.
Answer (D)
Note:
5C5, 6C5, 7C5 etc have unique values.
but 6C4 is the same as 6C2 - so if you were given that you can select r people out of 6 people in exactly 15 ways, r can take two distinct values.
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Karishma
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