permutation n combination

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permutation n combination

by rommysingh » Wed Aug 19, 2015 10:23 pm
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 Uva@90 » Thu Aug 20, 2015 8:29 am
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
Hi,

How you select Three woman from X women ==> XC3
and for men ==> YC2

We need to find XC3 * YC2

Stmt1 : (X+2)C3 = 56 ==> X = 6

Stmt 2 : X =Y+1

combine 1 and 2

X = 6 and y =5

So enough

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by Brent@GMATPrepNow » Thu Aug 20, 2015 8:38 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
Target question: How many different panels can be formed with these constraints?

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

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by Max@Math Revolution » Fri Aug 21, 2015 8:09 pm
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and equations ensures a solution.


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

==> In the original condition there are 2 variables (x,y) thus we need 2 equations to match the number of variables (This is essential in solving the problem). Both conditions (1) and (2) have 1 equations each, thus C is likely the answer and it actually turns out that C is the answer.

Going through the calculation, x+2C3=56 ==> (x+2)(x+1)x/3*2*1=56, and since x=6, y=5 the answer is C.



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by rommysingh » Sat Aug 22, 2015 7:22 am
Hi Uva is there a possible way of math u can show to reach x = 6 or is it the way brent has done that how u reached x = 6

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by Matt@VeritasPrep » Sun Aug 23, 2015 12:05 pm
This one has two steps: find the number of female groups and the number of male groups, then multiply the two together. (For instance, suppose I have two female groups and three male groups. They can be paired in 2 * 3 = 6 ways.)

To find the female groups, we do (x choose 3), or x! / (3! * (x-3)!), or (x * (x-1) * (x-2)) / 6. Similarly, the men are (y choose 2), or (y * (y-1)) / 2.

S1 tells us that ((x + 2) choose 3) = 56. This WILL give us a solution for x, but not for y. NOT SUFFICIENT

S2 tells us y in terms of x. So if we knew either x or y, we could now find the other. NOT SUFFICIENT

Together, S1 gives us x, and we plug that into S2 to find y. SUFFICIENT!

I think this conceptual approach is much quicker: the calculation in S1 is unnecessary, but is designed to get you to waste time if you don't think of the conceptual approach.

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by Matt@VeritasPrep » Sun Aug 23, 2015 12:13 pm
rommysingh wrote:Hi Uva is there a possible way of math u can show to reach x = 6 or is it the way brent has done that how u reached x = 6
It would look like this:

((x + 2) choose 3) = 56

(x + 2)! / (3! * (x + 2 -3)!) = 56

(x + 2)*(x + 1)*x / 3*2*1 = 56

(x + 2)*(x + 1)*x = 6*56 = 336

Now we have two options:

1:: Number properties

We want to factor 336 into three consecutive integers (x, x+1, and x+2). We notice that 336 divides by a lot of friendly numbers: 2, 3, 4, 6, 8, etc. 2*3*4 is too small, but 6*7*8 would work. Aha! So x = 6.

2:: Algebra

x³ + 3x² + 2x - 336 = 0

Since x is an integer solution, we know it must be a factor of 336. I'd then just use the number properties approach above, since the noninteger solutions will be too nasty to find without breaking this down to a quadratic using the integer root.

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by Max@Math Revolution » Sun Aug 23, 2015 12:24 pm
x+2C3=56, (x+2)(x+1)x/3*2*1=56, (x+2)(x+1)x=56*6=8*7*6 ==> x=6

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by gmatbeater1989 » Wed Oct 14, 2015 10:48 am
Max@Math Revolution wrote: ==> In the original condition there are 2 variables (x,y) thus we need 2 equations to match the number of variables (This is essential in solving the problem). Both conditions (1) and (2) have 1 equations each, thus C is likely the answer and it actually turns out that C is the answer.
That seems too easy. I don't think it always works. Anyone care to weigh in?

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by [email protected] » Wed Oct 14, 2015 2:50 pm
Hi gmatbeater1989,

You are correct - that type of thinking does not always lead to the correct answer. Data Sufficiency questions exist to test a number of different things, including organization, attention-to-detail, thoroughness of your thinking, etc. You will almost certainly see a few DS questions on Test Day in which the 'lazy thinker' would pick Answer C for the stated reason (and get the question wrong), when the correct answer is actually A, B or D.

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by Brent@GMATPrepNow » Thu Oct 15, 2015 5:50 am
I agree with Rich.

Here are some free videos related to this and other common misconceptions about Data Sufficiency questions.
Common GMAT Data Sufficiency Myths - Part I - https://www.gmatprepnow.com/module/gmat- ... video/1106
Common GMAT Data Sufficiency Myths - Part II - https://www.gmatprepnow.com/module/gmat- ... video/1107

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by hoppycat » Wed May 31, 2017 12:12 pm
[email protected] wrote:Hi gmatbeater1989,

You are correct - that type of thinking does not always lead to the correct answer. Data Sufficiency questions exist to test a number of different things, including organization, attention-to-detail, thoroughness of your thinking, etc. You will almost certainly see a few DS questions on Test Day in which the 'lazy thinker' would pick Answer C for the stated reason (and get the question wrong), when the correct answer is actually A, B or D.

GMAT assassins aren't born, they're made,
Rich
Are there times when we can use MathRev's rule? If so when?

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by [email protected] » Wed May 31, 2017 3:32 pm
Hi hoppycat,

While GMAT questions are often based on specific patterns - and you can take advantage of those patterns - you have to be careful when working through DS questions. Data Sufficiency prompts have no 'safety net', meaning that if you make a mistake, then there will be no natural way to 'catch' it (and you'll likely end up picking one of the wrong answers without knowing it). As such, if you spot what you think is a pattern in a DS question, you should still go through the process of proving what the correct answer is (without that proof, you might be making a silly mistake and losing out on some easy points).

Here's a simple example:

What is the value of 6X + 14Y?

Looking at this question, you might think... "Since I'm dealing with two variables, I need to know the values of X and Y - and that means that I'll probably need two distinct equations to answer this question".... Now take a good look at Fact 1....

1) 3X + 7Y = 12

With this equation, you do NOT need a second equation to answer the question. Simply 'doubling' this equation will get you the answer.

2(3X + 7Y) = 2(12)
6X + 14Y = 24

Thus, Fact 1 is sufficient on its own. If there was a second equation in Fact 2, then careless Test Taker might think that the correct answer was C (when it's clearly not). This type of question would be easy to miss if you were thinking in generalities. So again, whatever ideas you have when dealing with a DS question, make sure to put in the necessary work (on the PAD) to solve it.

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by Jay@ManhattanReview » Wed May 31, 2017 4:36 pm
[email protected] wrote:Hi hoppycat,

While GMAT questions are often based on specific patterns - and you can take advantage of those patterns - you have to be careful when working through DS questions. Data Sufficiency prompts have no 'safety net', meaning that if you make a mistake, then there will be no natural way to 'catch' it (and you'll likely end up picking one of the wrong answers without knowing it). As such, if you spot what you think is a pattern in a DS question, you should still go through the process of proving what the correct answer is (without that proof, you might be making a silly mistake and losing out on some easy points).

Here's a simple example:

What is the value of 6X + 14Y?

Looking at this question, you might think... "Since I'm dealing with two variables, I need to know the values of X and Y - and that means that I'll probably need two distinct equations to answer this question".... Now take a good look at Fact 1....

1) 3X + 7Y = 12

With this equation, you do NOT need a second equation to answer the question. Simply 'doubling' this equation will get you the answer.

2(3X + 7Y) = 2(12)
6X + 14Y = 24

Thus, Fact 1 is sufficient on its own. If there was a second equation in Fact 2, then careless Test Taker might think that the correct answer was C (when it's clearly not). This type of question would be easy to miss if you were thinking in generalities. So again, whatever ideas you have when dealing with a DS question, make sure to put in the necessary work (on the PAD) to solve it.

GMAT assassins aren't born, they're made,
Rich
Excellent example, Rich. I too thought of the exactly the same example. :)

-Jay
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by hoppycat » Sun Jun 04, 2017 7:37 am
thanks!