Mystery Teammates - Combinatorics

[karishma.bhargava]

Hello,

This question is from the Word Translations Question bank. The exact working on the question is as follows:

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.

I answered this question correctly but for the wrong reasons...When reading through the answer explanation- I don't understand how to get to the following equation for statement 1: 9! ÷ [(5!)(4!)] = 126; and the following equation for statement 2:8! ÷ [(5!)(3!)] = 56). Specifically I don't understand why you have to divide by 4! and 3!.

Thanks for your help in advance!

[RonPurewal]

Specifically I don't understand why you have to divide by 4! and 3!.

Thanks for your help in advance!

have you read the word translations strategy guide, and/or studied some sort of reference material on combinations?

this is the standard form of the formula for combinations. if you need to select r items from a total pool of n items, then the total number of ways in which you can make that selection is
n! / (r! (n - r)!)

in our strategy guide, this formula is also the general result of our ANAGRAM GRID method. if you don't feel like memorizing the formula itself, or you have difficulty doing so, then you can apply the "anagram grid" to each of the 2 problems.

note that the two numbers whose factorials appear in the denominator will add up to the number n. this happens above: 3 + 5 is indeed 8, and 4 + 5 is indeed 9.

--

in any case, you don't have to do any actual math to solve this problem.

it should be clear that the larger the pool, the more combinations that can be chosen.
the consequence is that if i tell you that there are a specific number of ways of choosing a group of some specific size, then you MUST know the size of the overall pool. for instance, if there are 126 ways of choosing 5 things out of a pool of n, then there can only be ONE value of n satisfying this criterion. we know this is the case, because any larger pool would give more than 126 such combinations, and any smaller pool would give fewer.

remember that, since this is data sufficiency, you don't care what the actual numbers are! just being able to find them, or just knowing that you can find them, is plenty good.

so, here's the minimal thought process required to solve this problem:

(1)
oh hey, this means i have a unique value for x + 2.
therefore, i have a unique value for x.
therefore, i can find how many combinations of 5 items i'd have.

(2)
oh hey, this means i have a unique value for x + 1.
therefore, i have a unique value for x.
therefore, i can find how many combinations of 5 items i'd have.

so, ans = (d)

[karishma.bhargava]

Thanks for your help! I need to go back and refresh my memory on that chapter! :)

[kinjal.nandy]

Heyy I have a doubt...
here they are asking how many different teams can be formed so we need to know the value of X to determine the value and for that we need both A abd B options to calculate the value...which comes out to be 7.
so I think the ans needs to be C.

Please let me know if I am incorrect

[esledge]

here they are asking how many different teams can be formed so we need to know the value of X to determine the value and for that we need both A abd B options to calculate the value...which comes out to be 7.
so I think the ans needs to be C.

Please let me know if I am incorrect

Your rephrased question "What is x?" is correct.

However, you are incorrect about which statement(s) provide the answer. The correct answer is D, either statement alone provides x.

Here is the math translation of each statement:
(1) (x+2)!/5!(x+2-5)! = 126 --> A little trial and error reveals that x+2=9, so x = 7.

(2) (x+1)!/5!(x+1-5)! = 56 --> A little trial and error reveals that x+1=8, so x = 7.

[yousuf_azim]

Dear Esledge,

Thanx for the ans.

BR

[tim]

On behalf of esledge, you’re welcome!

[numberonelabouchefan]

I have a conceptual question that was posed 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 total number of different 5-person teams would then be those summed together. What is wrong with interpreting the question this way????

[Sage Pearce-Higgins]

There's nothing wrong with interpreting the question in that way. Sure, it doesn't mention "single" teams. And you're right, if you have 11 people, then you could select two different teams. However, the question already includes that possibility - both those two teams are included in the list of possible teams. And you're over-complicating things by considering the limits to the second team. After all, we want to consider ALL the teams, not just those from the two arbitrary subgroups.

To make things clearer, take the example of selecting a 5-person team from the 11 people A B C D E F G H I J K . By your reasoning, we can pick 1 team A B C D E, leaving F G H I J K to form a number of other teams (6 possibilities, to be precise). However, we also want to consider possible teams comprised of, say, A B C and F and G. This shows that dividing the 11 people into subgroups of 5 and 6 wasn't useful; you'd actually end up with the same number of possibilities.

In terms of your Math, the formula you correctly apply - [(11*10*9*8*7)/(5!)] - incorporates all the choices, since your first choice is out of 11 people, then 10, etc. If you added on 2nd teams made up of those not chosen for the first team, you'd be double-counting some teams.

[MeekeyK670]

Here is are the equations you can derive from the statement and solve

Original Question needs to find x in xC5 = X!/(x-5)!5! --------------------(1)

Then from statement 1 we get

(x+2)C5 = 126 i.e (x+2)!/(x-3)!5! = 126------------------------(2)

From Statement 2 we get

(x+1)C3 = 56 i.e (x+1)!/(x-2)!3! =56 --------------------------(3)

Replace 2 with 3 and you get x=7

Hence 7C5 = 21

Happy learning!

[Sage Pearce-Higgins]

Well done MeekeyK670, those are correct equations. It's a nice challenge to actually solve a DS problem, but remember that you don't actually need to do this in a test situation. It's enough to see "I could make an equation that gives only one value for x, so that would be enough information to answer the question." Nice work though.