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Integers x and y are both positive, and x > y. How many different com

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Integers x and y are both positive, and x > y. How many different com  [#permalink]

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Integers x and y are both positive, and x > y. How many different committees of y people can be chosen from a group of x people?

(1) The number of different committees of x-y people that can be chosen from a group of x people is 3,060.

(2) The number of different ways to arrange x-y people in a line is 24.


Kudos for a correct solution.

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Re: Integers x and y are both positive, and x > y. How many different com  [#permalink]

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New post 07 May 2015, 09:37
Bunuel wrote:
Integers x and y are both positive, and x > y. How many different committees of y people can be chosen from a group of x people?

(1) The number of different committees of x-y people that can be chosen from a group of x people is 3,060.

(2) The number of different ways to arrange x-y people in a line is 24.


Kudos for a correct solution.


Choosing y people out of x = xCy = \(\frac{x!}{(y!)(x-y)!}\)

A) choosing x-y out of x = xC(x-y) = \(\frac{x!}{(x-x+y)!(x-y)!}\) = \(\frac{x!}{(y!)(x-y)!} = 3060\) --- Sufficient

B) No. of ways arranging x-y people in line = 24 => (x-y)! = 24 => x-y = 4, still no info about x and y in particular --- Insufficient

Hence Answer is A
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Integers x and y are both positive, and x > y. How many different com  [#permalink]

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New post 07 May 2015, 13:35
Bunuel wrote:
Kudos for a correct solution.


I'll give kudos to anyone who can post a correct solution without using any algebra. There is one.

Nice question Bunuel!
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Re: Integers x and y are both positive, and x > y. How many different com  [#permalink]

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New post 11 May 2015, 03:35
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Bunuel wrote:
Integers x and y are both positive, and x > y. How many different committees of y people can be chosen from a group of x people?

(1) The number of different committees of x-y people that can be chosen from a group of x people is 3,060.

(2) The number of different ways to arrange x-y people in a line is 24.


Kudos for a correct solution.


KAPLAN OFFICIAL SOLUTION:

The first step in this problem is to determine what we are really being asked. If we want to select committees of y people from a group of x people, we should use the combinations formula, which is n!/[k!/(n-k)!]. Remember, in this formula n is the number with which we start and k is the number we want in each group. Thus, we can reword the question as what does x!/[y!(x-y)!] equal?

Statement 1 tells us how many committees of x-y people we can make from our initial group of x people. If we plug this information into the combinations formula, we get x!/[(x-y)!(x-(x-y))!] = 3,060. This can be simplified to x!/[(x-y)!(x-x+y))!] = 3,060, which in turn is simplified to x!/[(x-y)!y!] = 3,060. The simplified equation matches the expression in our question, and gives us a numerical solution for it. Therefore, statement 1 is sufficient.

Statement 2 tells us how many ways we can arrange a number of people. The formula for arrangements is simply n!. In this case we have x-y people, thus (x-y)! = 24. Therefore, x-y must equal 4. However, we have no way of calculating what x and y actually are. This means that we cannot calculate the number of combinations in our question. Statement 2 is insufficient. So our final answer choice for this Data Sufficiency question is answer choice (A) or (1), Statement 1 is sufficient on its own, but Statement 2 is not.

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Re: Integers x and y are both positive, and x > y. How many different com  [#permalink]

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New post 19 Jul 2015, 12:54
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IanStewart wrote:
Bunuel wrote:
Kudos for a correct solution.


I'll give kudos to anyone who can post a correct solution without using any algebra. There is one.

Nice question Bunuel!



Correct me if I am wrong..
As 5c2=5c3 so likewise xcy=xc(x-y) and we have been given x>y. so from A we directly get the value.
B is Insufficient as we don't know anything about X or Y exact values.
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Re: Integers x and y are both positive, and x > y. How many different com  [#permalink]

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