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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- rommysingh
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Statement 1:rommysingh 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
No information about the number of men.
INSUFFICIENT.
Statement 2:
Since x and y can be any two consecutive nonnegative integers, INSUFFICIENT.
Statements combined:
Statement 1:
Since 56 = 7*8, at least 7 women are required to yield 56 different groups of 3.
TEST CASES:
From 7 women, the number of groups of 3 that can be formed = 7C3 = (7*6*5)/(3*2*1) = 35.
From 8 women, the number of groups of 3 that can be formed = 8C3 = (8*7*6)/(3*2*1) = 56.
From 9 women, the number of groups of 3 that can be formed = 9C3 = (9*8*7)/(3*2*1) = 84.
As indicated by the red case, 8 women are required to form 56 different groups of 3.
Since 8 is equal to two more than the ACTUAL number of women, x = 8-2 = 6.
Statement 2:
Since x=6 and x = y+1, y=5.
Since the values of x and y are known, it is possible to determine the number of 3-women, 2-men panels that can be formed.
SUFFICIENT.
The correct answer is C.
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- Brent@GMATPrepNow
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Target question: How many different panels can be formed with these constraints?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 ["x+2 CHOOSE 3 equals 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
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Using Statement 1
if two more women are availbe & number of ways of 3 groups are 56
x+2C3 = 56
(x+2)! / (3! * (x+2-3)! = 56
(x+2)(x+1)(x) (x-1)! / ( x-1)! = 56 * 6
(x+2)(x+1)(x) = 56 * 6
(x+2)(x+1)(x) = 6* 7 *8
x = 6
But Statement 1 gives only value of X i.e. number available women so Non Sufficient
Statement 2
x = y + 1
so y =5
now we can find 6C3 * 5C 2 Answer is C - Both statements together are sufficient.
if two more women are availbe & number of ways of 3 groups are 56
x+2C3 = 56
(x+2)! / (3! * (x+2-3)! = 56
(x+2)(x+1)(x) (x-1)! / ( x-1)! = 56 * 6
(x+2)(x+1)(x) = 56 * 6
(x+2)(x+1)(x) = 6* 7 *8
x = 6
But Statement 1 gives only value of X i.e. number available women so Non Sufficient
Statement 2
x = y + 1
so y =5
now we can find 6C3 * 5C 2 Answer is C - Both statements together are sufficient.