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Intern  Joined: 22 Dec 2009
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Question Stats: 88% (01:08) correct 12% (01:39) wrong based on 711 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!

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

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gmatJP wrote:
HI AbhayPrasanna.. Thanks for the reply...

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.

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

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2
We need a unique value.

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.
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Intern  Joined: 22 Dec 2009
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Re: How many different 5-person teams can be formed from a group  [#permalink]

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HI AbhayPrasanna.. Thanks for the reply...

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

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

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4
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.
Intern  Joined: 03 Dec 2010
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Re: How many different 5-person teams can be formed from a group  [#permalink]

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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
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Joined: 02 Sep 2009
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Re: How many different 5-person teams can be formed from a group  [#permalink]

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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 $$C^5_{(x+2)}=126$$.

Hope it's clear.
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Re: How many different 5-person teams can be formed from a group  [#permalink]

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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 $$C^5_{(x+2)}=126$$.

Hope it's clear.
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Re: How many different 5-person teams can be formed from a group  [#permalink]

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

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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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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on Combinations: math-combinatorics-87345.html

DS questions on Combinations: search.php?search_id=tag&tag_id=31
PS questions on Combinations: search.php?search_id=tag&tag_id=52

Tough and tricky questions on Combinations: hardest-area-questions-probability-and-combinations-101361.html

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

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gmatJP wrote:
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!

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 independent equations ensures a solution.

There are 1 variable x and 0 equation are given from the conditions of the original questions, so there is high chance (D) will be our answer.

The question asks what the value of $$_xC_2$$ is.

Condition (1)
$$_{x+2}C_5$$ = 126.
We can determine what the value of x is.
Hence, this condition is sufficient. We don't need calculate it.

Condition (2)
$$_{x+1}C_3$$ = 56.
We can determine what the value of x is.
Hence, this condition is sufficient. We don't need calculate it.

Therefore, the correct answer is D.

Keep in mind that DS is totally different from PS. DS questions do NOT ask what the exact value is.
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GMAT 1: 740 Q48 V44 GRE 1: Q165 V167 Re: How many different 5-person teams can be formed from a group  [#permalink]

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I have a conceptual question a student I tutor raised to me the other day. While we all assume that the question is asking for the number of ways a SINGLE 5-person team can be formed, the question never actually specifies this. In order words, if there were 11 people to choose from than we could actually make two different 5-person teams with one person left over. The number of combos for the 1st team[(11*10*9*8*7)/(5!)] would limit the number of combos for the 2nd team[(6*5*4*3*2)/(5!)]. The number of different 5-person teams would then be those summed together. What is wrong with interpreting the question this way????
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Re: How many different 5-person teams can be formed from a group  [#permalink]

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Bunuel wrote:
gmatJP wrote:
HI AbhayPrasanna.. Thanks for the reply...

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.

Hope it helps.

Hi Bunuel!

$$\frac{5!}{x!(5-x)!}=10$$ --> $$x!(5-x)!=12$$ --> $$x=3$$ or $$x=2$$.

In this how does x!(5-x)! give us x as 3 or 2? what is the calculation like? Re: How many different 5-person teams can be formed from a group   [#permalink] 07 Dec 2019, 00:08
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