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Integers x and y are both positive, and x > y. How many different com
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07 May 2015, 03:18
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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 xy people that can be chosen from a group of x people is 3,060. (2) The number of different ways to arrange xy 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
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07 May 2015, 10:37
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 xy people that can be chosen from a group of x people is 3,060.
(2) The number of different ways to arrange xy people in a line is 24.
Kudos for a correct solution. Choosing y people out of x = xCy = \(\frac{x!}{(y!)(xy)!}\) A) choosing xy out of x = xC(xy) = \(\frac{x!}{(xx+y)!(xy)!}\) = \(\frac{x!}{(y!)(xy)!} = 3060\)  SufficientB) No. of ways arranging xy people in line = 24 => (xy)! = 24 => xy = 4, still no info about x and y in particular  InsufficientHence Answer is A



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Integers x and y are both positive, and x > y. How many different com
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07 May 2015, 14:35
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
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11 May 2015, 04:35
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 xy people that can be chosen from a group of x people is 3,060.
(2) The number of different ways to arrange xy 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!/(nk)!]. 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!(xy)!] equal? Statement 1 tells us how many committees of xy people we can make from our initial group of x people. If we plug this information into the combinations formula, we get x!/[(xy)!(x(xy))!] = 3,060. This can be simplified to x!/[(xy)!(xx+y))!] = 3,060, which in turn is simplified to x!/[(xy)!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 xy people, thus (xy)! = 24. Therefore, xy 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. ~Bret Ruber
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Re: Integers x and y are both positive, and x > y. How many different com
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19 Jul 2015, 13:54
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(xy) 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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