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From a group of 8 people, it is possible to create 56
[#permalink]
18 Aug 2014, 10:24

1

Expert Reply

4

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Question Stats:

From a group of 8 people, it is possible to create 56 different k-person committees. Which of the following could be the value of k ?

Indicate all such values.

A)1

B)2

C)3

D)4

E)5

F)6

G)7

Indicate all such values.

A)1

B)2

C)3

D)4

E)5

F)6

G)7

Re: GRE Math Challenge #4-From a group of 8 people
[#permalink]
03 Sep 2018, 18:00

Can someone please explain the answer?

Re: GRE Math Challenge #4-From a group of 8 people
[#permalink]
03 Sep 2018, 18:31

CE is correct

Mathematically, 8 choose k = 56, therefore k=2 or 8. (Check out the formula for permutation)

You could also plug in each choice below and validate it. When you're picking 2 people out of 8, the first time you have 8 choices, second you have 7, so total 8*7=56,

6 also applies because picking 6 out of 8 people for the committee is the same as picking 2 out of 8 people NOT for the committee.

The other options don't make sense.

_________________

NixonDutt wrote:

Can someone please explain the answer?

Mathematically, 8 choose k = 56, therefore k=2 or 8. (Check out the formula for permutation)

You could also plug in each choice below and validate it. When you're picking 2 people out of 8, the first time you have 8 choices, second you have 7, so total 8*7=56,

6 also applies because picking 6 out of 8 people for the committee is the same as picking 2 out of 8 people NOT for the committee.

The other options don't make sense.

_________________

Re: GRE Math Challenge #4-From a group of 8 people
[#permalink]
14 Oct 2018, 17:22

1

To chose k people from 8 to form 56 k-person committees:

8!/x!(8-x)!= 56

I chose at random with 3, so 8!/3!5! = 8*7*6*5!/(3*2)5! = 8*7*6/6 = 56

So 3 and 5 are the answers because even when you switch 3! for 5! in the first half of the denominator, you still get the same value in the end.

I hope that makes sense!

8!/x!(8-x)!= 56

I chose at random with 3, so 8!/3!5! = 8*7*6*5!/(3*2)5! = 8*7*6/6 = 56

So 3 and 5 are the answers because even when you switch 3! for 5! in the first half of the denominator, you still get the same value in the end.

I hope that makes sense!

From a group of 8 people, it is possible to create exactly 5
[#permalink]
11 Dec 2018, 14:17

Expert Reply

From a group of 8 people, it is possible to create exactly 56 different k-person committees. Which of the following could be the value of k?

Indicate all such values.

A. 1

B. 2

C. 3

D. 4

E. 5

F. 6

G. 7

_________________

Indicate all such values.

A. 1

B. 2

C. 3

D. 4

E. 5

F. 6

G. 7

_________________

Re: From a group of 8 people, it is possible to create exactly 5
[#permalink]
12 Dec 2018, 09:36

4

Carcass wrote:

From a group of 8 people, it is possible to create exactly 56 different k-person committees. Which of the following could be the value of k?

Indicate all such values.

A. 1

B. 2

C. 3

D. 4

E. 5

F. 6

G. 7

Indicate all such values.

A. 1

B. 2

C. 3

D. 4

E. 5

F. 6

G. 7

Explanation::

As, 56 different k - committee can be formed by the group of 8 people, that means

8Ck = 56

Now it is require to check the value of k from the option which results to 56

When K= 5,

\(8C5 = \frac{(8*7*6*5!)}{(5!*3!)} = 56\)

When K= 3,

\(8C3 = \frac{(8*7*6*5!)}{(5!*3!)} = 56\)

_________________

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Re: From a group of 8 people, it is possible to create exactly 5
[#permalink]
09 Jan 2019, 01:48

1

Expert Reply

we know this is Combination based, and order doesnt matter, so we have

8cK=56

8!/(k!*(8-k!)=56

lets ditch the fraction

8!= 56(k!*8-k)!

divide 8! by 56 (which is 8*7) and we have

6! =(k!*8-k)!

720 =(k!*8-k)!

Dont bother plugging in 1, since that will give you a multiple of 7, and that cant make 720, try pluggin in 2

720=(2!*6!) ===nope 2!*6!= 1440

work your way up

720=(3!*5!) that works

try k=4

720= 4!*4!.. nope, that equals 576

try k =5

720=5!*3!, which is identical to k=3

try k=6

720= 6!*2!... nope

you cant go higher than that, it wont make sense.

The answers are C and E

8cK=56

8!/(k!*(8-k!)=56

lets ditch the fraction

8!= 56(k!*8-k)!

divide 8! by 56 (which is 8*7) and we have

6! =(k!*8-k)!

720 =(k!*8-k)!

Dont bother plugging in 1, since that will give you a multiple of 7, and that cant make 720, try pluggin in 2

720=(2!*6!) ===nope 2!*6!= 1440

work your way up

720=(3!*5!) that works

try k=4

720= 4!*4!.. nope, that equals 576

try k =5

720=5!*3!, which is identical to k=3

try k=6

720= 6!*2!... nope

you cant go higher than that, it wont make sense.

The answers are C and E

Re: From a group of 8 people, it is possible to create exactly 5
[#permalink]
09 Jan 2019, 06:20

Expert Reply

Carcass wrote:

From a group of 8 people, it is possible to create exactly 56 different k-person committees. Which of the following could be the value of k?

Indicate all such values.

A. 1

B. 2

C. 3

D. 4

E. 5

F. 6

G. 7

Indicate all such values.

A. 1

B. 2

C. 3

D. 4

E. 5

F. 6

G. 7

So, 8Ck = 56 = 8*7...

You can substitute k as 2 and see.. 8C2=\(\frac{8*7}{2}\), so take k as 3

8C3=\(\frac{8*7*6}{3!}=56\), so 3 is one value..

Also 8Ck = 8C(8-k).....if k is 3, 8-k=8-3=5.. Thus 3 and 5 are the answers.

otherwise \(8Ck=56...\frac{8*7*6!}{(8-k)!k!}=56.....\frac{6!}{(8-k)!k!}=1....6!=(8-k)!k!....6*5!=(8-k)!k!....3!5!=(8-k)!k!\)

Thus, k can be 3 or 5.

_________________

Some useful Theory.

1. Arithmetic and Geometric progressions : https://gre.myprepclub.com/forum/progressions-arithmetic-geometric-and-harmonic-11574.html#p27048

2. Effect of Arithmetic Operations on fraction : https://gre.myprepclub.com/forum/effects-of-arithmetic-operations-on-fractions-11573.html?sid=d570445335a783891cd4d48a17db9825

3. Remainders : https://gre.myprepclub.com/forum/remainders-what-you-should-know-11524.html

4. Number properties : https://gre.myprepclub.com/forum/number-property-all-you-require-11518.html

5. Absolute Modulus and Inequalities : https://gre.myprepclub.com/forum/absolute-modulus-a-better-understanding-11281.html

1. Arithmetic and Geometric progressions : https://gre.myprepclub.com/forum/progressions-arithmetic-geometric-and-harmonic-11574.html#p27048

2. Effect of Arithmetic Operations on fraction : https://gre.myprepclub.com/forum/effects-of-arithmetic-operations-on-fractions-11573.html?sid=d570445335a783891cd4d48a17db9825

3. Remainders : https://gre.myprepclub.com/forum/remainders-what-you-should-know-11524.html

4. Number properties : https://gre.myprepclub.com/forum/number-property-all-you-require-11518.html

5. Absolute Modulus and Inequalities : https://gre.myprepclub.com/forum/absolute-modulus-a-better-understanding-11281.html

Re: From a group of 8 people, it is possible to create 56
[#permalink]
16 Dec 2019, 12:12

Expert Reply

Bump for further discussion

_________________

_________________

Re: From a group of 8 people, it is possible to create exactly 5
[#permalink]
01 May 2020, 02:54

One important doubt here, it says 56 different groups, but it does not say that there CAN ONLY BE 56 groups. So even if we take 4 people out of 8 (where it's the maximum number of configurations possible), we CAN TAKE 56 different groups from this list. So shouldn't 4 be included as well?

Re: From a group of 8 people, it is possible to create exactly 5
[#permalink]
01 May 2020, 12:39

2

Zohair123 wrote:

One important doubt here, it says 56 different groups, but it does not say that there CAN ONLY BE 56 groups. So even if we take 4 people out of 8 (where it's the maximum number of configurations possible), we CAN TAKE 56 different groups from this list. So shouldn't 4 be included as well?

Great question!!

Many students have posed very similar questions.

For example, if a question tells us that A woman owns 5 dogs, must we assume that she has exactly 5 dogs?

After all, she could have 6 dogs, since it would still be true that there are 5 dogs in her possession (plus 1 more).

If this were the accepted standard, it would be next to impossible to phrase questions that are free from ambiguity.

So, on the GRE, if you're told that there are X things, we can assume that there are exactly X things.

So, for the question above, we can assume that there are exactly 56 different k-person committees possible.

If the author intended to phrase the question as you are suggesting, it would read something like "From a group of 8 people, we can create AT LEAST 56 different k-person committees. Which of the following could be the value of k?"

Cheers,

Brent

_________________

Re: From a group of 8 people, it is possible to create exactly 5
[#permalink]
11 Aug 2023, 19:37

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