Which of the following is an integer?

Question

I.  \frac{12!}{6!}

II.  \frac{12!}{8!}

III.  \frac{12!}{7!5!}

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I, II, and III

ScottTargetTestPrep · Accepted Answer

Before actually solving this problem, let's review how factorials can be expanded and expressed. As as example, we can use 5!.
 
5! could be expressed as:
 
5! 
 
5 x 4! 
 
5 x 4 x 3! 
 
5 x 4 x 3 x 2! 
 
5 x 4 x 3 x 2 x 1!
 
Understanding how this factorial expansion works will help us work our way through each answer choice, especially answer choices 1 and 2.
 
I. 12!/6! 
 
Since we know that factorials can be expanded, we now know that:
 
12! = 12 x 11 x 10 x 9 x 8 x 7 x 6!
 
Plugging this in for answer choice 1, we have:
 
(12 x 11 x 10 x 9 x 8 x 7 x 6!)/6! = 12 x 11 x 10 x 9 x 8 x 7, which is an integer.
 
II. 12!/8! 
 
Once again, since we know that factorials can be expanded, we now know that:
 
12! = 12 x 11 x 10 x 9 x 8!
 
Plugging this in for answer choice 2, we have:
 
(12 x 11 x 10 x 9 x 8!)/8! = 12 x 11 x 10 x 9, which is an integer.
 
III. 12!/(7!5!) 
 
Once again, since we know that factorials can be expanded, we now know that:
 
12! = 12 x 11 x 10 x 9 x 8 x 7!
 
Plugging this in for answer choice 3 gives us:
 
(12 x 11 x 10 x 9 x 8 x 7!)/(7!5!) 
 
(12 x 11 x 10 x 9 x 8)/(5 x 4 x 3 x 2 x 1)
 
(12 x 11 x 10 x 9 x 8)/(12 x 10 x 1)
 
11 x 9 x 8, which is an integer.
 
We see that the quantities in Roman numerals I, II and III are all integers.

Alternate solution: 
 
For any positive integers m, n and p,
 
1) If m > n, then m!/n! is always an integer.
 
2) If m = n + p, then m!/(n!p!) is always an integer (which is in fact mCp).
 
From the above two facts, we see that all three quotients in the Roman numerals must be integers.

Answer: E

EMPOWERgmatRichC · Answer

Hi All,

Beyond the arithmetic that sdrandom1 pointed out, these calculations fall into some patterns that you are likely to see (in some form) on Test Day.

12!/6!

When dealing with INDIVIDUAL factorials, if the factorial in the numerator is GREATER than the factorial in the denominator, then you will end up with an integer. Here, since 12 > 6, 12!/6! will simplify to an integer.

12!/7!5!

You might not have studied the Combination Formula yet, but this calculation is what you would end up with when answering the question "how many different combinations of 7 people can you form from a group of 12 people?" In these types of "Combination" calculations, the number of groups will always be an integer, so 12!/7!5! will simplify to an integer.

Final Answer:  E

GMAT assassins aren't born, they're made, 
Rich

alwynjoseph · Answer

To me the question looks straight forward.
we know 12! is multiple of all numbers from 1 to 12, so dividing it by 6! or 8! is pretty much sure an integer.

now coming to the 3rd options 12! by 5! leaves 12*11*10*9*8*7*6/1*2*3*4*5*6*7 which results in integer, so the answe E is right

sdrandom1 · Answer

There is no shortcut persay...

Clear E.

12!/6! = 12*11*10*9*8*7*6!/6! = Integer
12!/8! = 12*11*10*9*8!/8! = Integer
12!/7!*5! = 12*11*10*9*8/5*4*3*2*1 = integer

So all 3 are correct.

IanStewart · Answer

Every factorial is divisible by every smaller factorial, so 12!/6! and 12!/8! are integers. 12!/(7!*5!) is the number of ways to choose 7 things from a group of 12 if order is irrelevant, so it must also be an integer.

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thangvietname · Answer

case 1 and case 2 are simple but case 3 is harder. for case 3, we have to write down all the factors.

energetics · Answer

For III, I went through and crossed out everything that's reduced to visualize...
12*11* 10 2* 9 3* 8*7*6*5*4*3*2*1  /  7*6*5*4*3*2*1  *  5*4*3*2*1

So you're left with 12*11*2*3 (from taking 5 and 3 from 10 and 9) = 792. Makes sense.

ScottTargetTestPrep  
As a rule, you said m = n + p, then m!/(n!p!). When does the factorial in the denominator make the result not an integer? Am I right to think this has to do with the number of prime factors?
Because 12!/8!5! and 12!/9!5! are also integers, while 12!/10!5! isn't. If we had 11*2^3*3^2 left with 7! on the bottom then 8! would be 2^3 and 9! would be 3^2 both of which we can still reduce... while 10! gives us a factor of 5 which we don't have.

ScottTargetTestPrep · Answer

So you ask a really great qestion. So, as you've shown above, when m < n + p, there is no hard and fast rule to determien when m!/(n!p!) is an integer and when it is not.

RobertVericPrep · Answer

This question gives you a great chance to review factorials, and once you answer this question you'll be well-prepared for other questions that involve factorials.

Here's an explanation on YouTube

https://www.youtube.com/watch?v=U7kymGI ... PfYZOmjn_0

Answer (E)

bumpbot · Answer

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Which of the following is an integer?

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Which of the following is an integer?

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