Can someone please explain the answer?

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.

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!

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

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

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.

Bump for further discussion

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

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