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How many different 5-person teams can be formed from a group [#permalink]

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22 Jun 2010, 19:14

11

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A

B

C

D

E

Difficulty:

5% (low)

Question Stats:

89% (01:46) correct
11% (01:07) wrong based on 520 sessions

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How many different 5-person teams can be formed from a group of x individuals?

(1) If there had been x + 2 individuals in the group, exactly 126 different 5-person teams could have been formed.

(2) If there had been x + 1 individuals in the group, exactly 56 different 3-person teams could have been formed.

The answer is D and I know how to figure out now but is there any trick to know each question is sufficient without actual compute? cuz its time consuming until I found out e.g. 1) is 9!/5!4!

I understood the first half but I dont follow your explanation of 'the only non-unique solution will be for the "r" in nCr eg. (8C3 = 8C5) but for n not equal to m, nCr can never be equal to mCr' ...

How do you know (x+2) C 5 = 126 is computable

praveenism wrote:

I agree with GmatJP...but I think we need to assume that whatever is given in the options is assumed to be true.So the x has to have an integral value. But my concern is..what if the polynomial equation so formed of degree 5 have multiple solutions. In that case, the options fails to answer the situation.

@Abhayprasanna : The logic is correct and infact I also choose D going by the same logic.

The point here is the following: Suppose we are told that there are 10 ways to choose \(x\) people out of 5. What is \(x\)? \(C^x_5=10\) --> \(\frac{5!}{x!(5-x)!}=10\) --> \(x!(5-x)!=12\) --> \(x=3\) or \(x=2\). So we cannot determine single numerical value of \(x\). Note that in some cases we'll be able to find \(x\), as there will be only one solution for it, but generally when we are told that there are \(n\) ways to choose \(x\) out of \(m\) there will be (in most cases) two solutions of \(x\) possible.

But if we are told that there are 10 ways to choose 2 out of \(x\), then there will be only one value of \(x\) possible --> \(C^2_x=10\) --> \(\frac{x!}{2!(x-2)!}=10\) --> \(\frac{x(x-1)}{2!}=10\) --> \(x(x-1)=20\) --> \(x=5\).

In our original question, statement (1) says that there are 126 ways to choose 5 out of \(x+2\) --> there will be only one value possible for \(x+2\), so we can find \(x\). Sufficient.

Just to show how it can be done: \(C^5_{(x+2)}=126\) --> \((x-2)(x-1)x(x+1)(x+2)=5!*126=120*126=(8*5*3)*(9*7*2)=5*6*7*8*9\) --> \(x=7\). Basically we have that the product of five consecutive integers (\((x-2)(x-1)x(x+1)(x+2)\)) equal to some number (\(5!*126\)) --> only one such sequence is possible, hence even though we have the equation of 5th degree it will have only one positive integer solution.

Statement (2) says that there are 56 ways to choose 3 out of \(x+1\) --> there will be only one value possible for \(x+1\), so we can find \(x\). Sufficient.

\(C^3_{(x+1)}=56\) --> \((x-1)x(x+1)=3!*56=6*7*8\) --> \(x=7\). Again we have that the product of three consecutive integers (\((x-1)x(x+1)\)) equal to some number (\(3!*56\)) --> only one such sequence is possible, hence even though we have the equation of 3rd degree it will have only one positive integer solution.

1. (x+2) C 5 = 126. There is only one possible value for x+2 that would yield a value of 126. Don't bother trying to find out what it is. Remember, the only non-unique solution will be for the "r" in nCr eg. (8C3 = 8C5) but for n not equal to m, nCr can never be equal to mCr.

2. (x+1) C 3 = 56 Again, you should be able to see that there can be only one value of x+1 that would yield a value of 56. Why bother finding out what the value is? As long as we have an equation in one variable, we can find a value.

Bunuel supplied an awesome and exhaustive mathematical algebraic explanation. Perhaps it will be of benefit to review the concept verbally as well.

8C5 = 8C3 because everytime we pull a subgroup of 5 objects from the bigger group of 8, we can see that we are also "setting aside" a subgroup of 3. Likewise, everytime we pull out a subgroup of 3, we also set aside a subgrup of 5. So, the number of ways we can pull out 5 object subgroups must be the same as the number of ways we can pull out 3 object subgroups.

But if there are 126 ways to pull 5 objects from a big group "x + 2", then "x+2" must be just one value. If it were not, then it would imply that increasing or decresing the size of the big group doesn't necessarily affect how many ways you can pull out a smaller subgroup--surely an absurd conclusion. Absurd because clearly there are more ways to pull 5 objects out of a set of 100 than out of a set of, say, 10.

I understood the first half but I dont follow your explanation of 'the only non-unique solution will be for the "r" in nCr eg. (8C3 = 8C5) but for n not equal to m, nCr can never be equal to mCr' ...

I agree with GmatJP...but I think we need to assume that whatever is given in the options is assumed to be true.So the x has to have an integral value. But my concern is..what if the polynomial equation so formed of degree 5 have multiple solutions. In that case, the options fails to answer the situation.

@Abhayprasanna : The logic is correct and infact I also choose D going by the same logic.
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Can any1 pls explain the calculation in a bit more simplier way. I didn't get how,

3Cx+1 =56 and how, 5Cx+2 = 126 ?

Thnx

# of ways to pick \(k\) objects out of \(n\) distinct objects is \(C^k_n=\frac{n!}{(n-k)!*k!}\).

# of ways to pick 3 people out of x+1 people is \(C^3_{(x+1)}=\frac{(x+1)!}{(x+1-3)!*3!}=\frac{(x+1)!}{(x-2)!*3!}\). Now, since \((x+1)!=(x-2)!*(x-1)*x*(x+1)\) then \((x-2)!\) get reduced and we'll have: \(\frac{(x-1)*x*(x+1)}{3!}\). We are told that this equals to 56: \(\frac{(x-1)*x*(x+1)}{3!}=56\) --> \((x-1)x(x+1)=3!*56=6*7*8\) --> \(x=7\).

You can apply similar logic to \(C^5_{(x+2)}=126\).

Re: How many different 5-person teams can be formed from a group [#permalink]

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20 Aug 2012, 03:39

Bunuel,

Can you please confirm if my understanding is correct?

(x+1)!= 3 consecutive numbers (x-1) x (x+1) (x+3)!= Basically 7 consecutive numbers i.e. (x-3) (x-2) (x-1) x (x+1) (x+2) (x+3) (x+2)!= 5 consecutive numbers i.e. (x-2) (x-1) x (x+1) (x+2)

Bunuel wrote:

priyalr wrote:

Hello to all,

Can any1 pls explain the calculation in a bit more simplier way. I didn't get how,

3Cx+1 =56 and how, 5Cx+2 = 126 ?

Thnx

# of ways to pick \(k\) objects out of \(n\) distinct objects is \(C^k_n=\frac{n!}{(n-k)!*k!}\).

# of ways to pick 3 people out of x+1 people is \(C^3_{(x+1)}=\frac{(x+1)!}{(x+1-3)!*3!}=\frac{(x+1)!}{(x-2)!*3!}\). Now, since \((x+1)!=(x-2)!*(x-1)*x*(x+1)\) then \((x-2)!\) get reduced and we'll have: \(\frac{(x-1)*x*(x+1)}{3!}\). We are told that this equals to 56: \(\frac{(x-1)*x*(x+1)}{3!}=56\) --> \((x-1)x(x+1)=3!*56=6*7*8\) --> \(x=7\).

You can apply similar logic to \(C^5_{(x+2)}=126\).

Re: How many different 5-person teams can be formed from a group [#permalink]

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22 Aug 2012, 14:23

manjeet1972 wrote:

Bunuel,

Can you please confirm if my understanding is correct?

(x+1)!= 3 consecutive numbers (x-1) x (x+1) (x+3)!= Basically 7 consecutive numbers i.e. (x-3) (x-2) (x-1) x (x+1) (x+2) (x+3) (x+2)!= 5 consecutive numbers i.e. (x-2) (x-1) x (x+1) (x+2)

[/quote]

Plug in numbers yourself. let x = 4, so you have (4+1)! = 5! = 5 * 4 * 3 * 2 * 1. But if x = 219739218731927 then you have a lot more terms.

Instead, try to compare things logically. Say you have (x+1)! / (x-4)! ...now what happens normally for simplifying factorials? If you have 11!/6! then it's 11 * 10 * 9 * 8 * 7. You go down by 1 term until you hit the denominator and remove. Same concept.

So now (x+1)! = (x+1) * x * (x-1) * (x-2) * (x-3) * (x-4) ...until you reach the end. But since we know the denominator is (x-4)! then we only have to go up to (x+1) * x * (x-1) * (x-2) * (x-3) on the numerator. Once you simplify the factorials, you can line them up like how Bunuel did and match it with the expanded form of (5!)(126) which would take a while to flat out compute but you just need to determine if it's solvable.

Re: How many different 5-person teams can be formed from a group [#permalink]

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22 Aug 2012, 22:38

Good explanation. Through practice I will learn.

Injuin wrote:

manjeet1972 wrote:

Bunuel,

Can you please confirm if my understanding is correct?

(x+1)!= 3 consecutive numbers (x-1) x (x+1) (x+3)!= Basically 7 consecutive numbers i.e. (x-3) (x-2) (x-1) x (x+1) (x+2) (x+3) (x+2)!= 5 consecutive numbers i.e. (x-2) (x-1) x (x+1) (x+2)

Plug in numbers yourself. let x = 4, so you have (4+1)! = 5! = 5 * 4 * 3 * 2 * 1. But if x = 219739218731927 then you have a lot more terms.

Instead, try to compare things logically. Say you have (x+1)! / (x-4)! ...now what happens normally for simplifying factorials? If you have 11!/6! then it's 11 * 10 * 9 * 8 * 7. You go down by 1 term until you hit the denominator and remove. Same concept.

So now (x+1)! = (x+1) * x * (x-1) * (x-2) * (x-3) * (x-4) ...until you reach the end. But since we know the denominator is (x-4)! then we only have to go up to (x+1) * x * (x-1) * (x-2) * (x-3) on the numerator. Once you simplify the factorials, you can line them up like how Bunuel did and match it with the expanded form of (5!)(126) which would take a while to flat out compute but you just need to determine if it's solvable.[/quote]

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