Question Bank - Word Translations Q16

[SaoCX]

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.

Source: MGMAT Question Bank - Word Translations Q16

How this can be solved WITHOUT testing values?

Thanks for your help.

[givemeanid]

You are asked the value of xC5; this means you need the value of x.

1. (x+2)C5 = 126. You can calculate x.
SUFFICIENT.

2. (x+1)C3 = 56. You can calculate x.
SUFFICIENT.

Answer is D.

[Luci]

In this problem the statements do not contradict each other but they dont yield the same resutl, do they?

For statement 1 we have: If there had been x + 2 individuals in the group, exactly 126 different 5-person teams could have been formed.
Exactly 126 different 5-person teams= 126*5=630 people
If we subtract 2, then x=628

For statement 2 we have: If there had been x + 1 individuals in the group, exactly 56 different 3-person teams could have been formed.
exactly 56 different 3-person teams= 56*3=168
If we add 1, then x=169

How is this posible? We can solve for x but we obtain different results. I´m lost here

Luci

[givemeanid]

For statement 1 we have: If there had been x + 2 individuals in the group, exactly 126 different 5-person teams could have been formed.
Exactly 126 different 5-person teams= 126*5=630 people
If we subtract 2, then x=628

For statement 2 we have: If there had been x + 1 individuals in the group, exactly 56 different 3-person teams could have been formed.
exactly 56 different 3-person teams= 56*3=168
If we add 1, then x=169

Luci, nCr = n!/(r! * (n-r)!)
Exactly 126 different 5 person teams DOES NOT mean 126*5 people. It means, how many different combinations are possible when people are grouped together in 5s.

(x+2)C5 = 126
You can calculated x+2 = 9 (9C5 = 9!/(5!*4!) = 126)
x = 7


From the second statement, (x+1)C3 = 56
8C3 = 56
x+1 = 8
x = 7

Both statements are consistent with each other.

[dbernst]

givemeanid, thanks again for the help!

-dan

[unique]

givemeanid,

I must be a real nut. Please explain.

I understand the combinations part. But how did you pick x + 2 = 9 for the first solution 9C5 = 126

and how did you pick x+1 = 8 for the second problem.

[givemeanid]

givemeanid,

I must be a real nut. Please explain.

I understand the combinations part. But how did you pick x + 2 = 9 for the first solution 9C5 = 126

and how did you pick x+1 = 8 for the second problem.

unique, remember that you DO NOT need to do the calculations. You just need to know that you can calculate 'x' from that equation and move on.

Regarding the actual calculations, I just picked up a few numbers and did a brute force. Just looking at the equation (x+2)C5 = 126, I knew x+2 had to be 8 or 9. To really do it the right way, you have expand factorials and that would defeat the purpose of GMAT questions. From what I've gathered over the past few days of my prep, if you are bogged down with calculations, you are missing a key piece of GMAT jigsaw.

But again, remember, you do not need to do calcs.

[JadranLee]

Good point, givemeanid.

-Jad