hard DS problem - can someone help?

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hard DS problem - can someone help?

by mikepamlyla » Sun Feb 02, 2014 11:10 am
A certain panel is to be composed of exactly three women and exactly two men, chosen from x women and y men. How many different panels can be formed with these constraints?

(1) If two more women were available for selection, exactly 56 different groups of three women could be selected.

(2) x = y + 1

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by Brent@GMATPrepNow » Sun Feb 02, 2014 12:02 pm
mikepamlyla wrote:A certain panel is to be composed of exactly three women and exactly two men, chosen from x women and y men. How many different panels can be formed with these constraints?

(1) If two more women were available for selection, exactly 56 different groups of three women could be selected.

(2) x = y + 1


First recognize that, since the order of the selected people does not matter, we can use combinations to solve this. We can select 3 women from x women in xC3 ways, and we can select 2 men from y men in yC2 ways. So, the total number of possible panels = (xC3)(yC2)

As you can see,the answer to the target question will depend solely on the individual values of x and y.

Statement 1: If two more women were available for selection, exactly 56 different groups of three women could be selected.
Since there's no information about the number of men, statement 1 is NOT SUFFICIENT

Statement 2: x = y + 1
There are several pairs that meet this condition. Here are two:
Case a: x = 3 and y = 2, in which case there's only 1 possible panel (since we'd have no choice but to select all 5 people)
Case b: x = 201 and y = 200, in which case there are TONS of possible panels
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 basically tells us that (x+2)C3 = 56
Do we need to solve for x? NO. We need only recognize that we COULD solve for x.
Let's start checking a few possible values.
3C3 = 1
4C3 = 4
5C3 = 10
6C3 = 20

Aside: if anyone is interested, we have a free video on calculating combinations (like 6C3) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

We can see that the numbers keep increasing, so there will be ONLY ONE value of x such that (x+2)C3 = 56 [incidentally, 8C3 = 56. So, (x+2) = 8, which means x = 6. Of course, that doesn't really matter since we need only recognize that we COULD determine the value of x]
So, from statement 1, we COULD determine the value of x
Once we know the value of x, we can use statement 2 to determine the value of y.
At this point, we can answer the target question with certainty, so the combined statements are SUFFICIENT

Answer = C

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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