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

Question

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 .

Bunuel · Accepted Answer

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!/ . 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!/  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!/  = 3,060. This can be simplified to x!/  = 3,060, which in turn is simplified to x!/  = 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.

~Bret Ruber

King407 · Answer

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

IanStewart · Answer

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

Nice question Bunuel!

sahil7389 · Answer

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.

pudu · Answer

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?
xCy=x!/y!*(x-y) ?

(1) The number of different committees of x-y people that can be chosen from a group of x people is 3,060.
xCx-y=3060
x!/y!*(x-y)!= 3060 . Sufficient

(2) The number of different ways to arrange x-y people in a line is 24.
(x-y)!=24. No idea about x! and y!. So insufficient.

Hence A.

DrAnkita91 · Answer

x!/y!*(x-y)!= 3060
can please anyone solve this further
thanking you in advance

bumpbot · Answer

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

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