**Learn how to simplify these seemingly devilishly complicated GMAT Quant problems!**

First, consider these problems

3) Consider these three quantities

Rank these three quantities from least to greatest.

- (A) I, II, III

- (B) I, III, II

- (C) II, I, III

- (D) II, III, I

- (E) III, I, II

These are challenging problems, especially the third one! With a few simple insights about factorials, though, you will be able to manage all of these.

## The factorial

First of all, a few basics. The factorial is a function we can perform on any positive integer. The expression 5! (read “five factorial”) means the product of all the positive integers from that number down to one. In this particular case: 5! = 5*4*3*2*1 = 120. Here are the first ten factorials, just to give you a sense:

1! = 1

2! = 2*1 = 2

3! = 3*2*1 = 6

4! = 4*3*2*1 = 24

5! = 5*4*3*2*1 = 120

6! = 6*5*4*3*2*1 = 720

7! = 7*6*5*4*3*2*1 = 5,040

8! = 8*7*6*5*4*3*2*1 = 40,320

9! = 9*8*7*6*5*4*3*2*1 = 362,880

10! = 10*9*8*7*6*5*4*3*2*1 = 3,628,800

It’s a good idea to have the first five memorized, simply because those come up frequently on the GMAT —the first five are pretty easy to figure out on your own anyway. *Nobody* expects you to have the last five here memorized. I give them here purely to give you a sense of how quickly the factorials grow. By the time we get to (10!), we are already over a million! Similarly, (13!) is more than a billion, and (15!) is more than a trillion. Holy schnikes!

## Operations with factorials

Here are some big ideas to help you when you have to perform arithmetic operations with factorials.

**Big Idea #1: every factorial is a factor of every higher factorial**

The number 73! must be divisible by 72!, by 47!, by 12!, etc. The number 73! automatically has at least 72 known factors — all the factorials less than it!

**Big Idea #2: you can “unpack” one factorial down to another. **

Think about 8! —- we know:

8! = 8*7*6*5*4*3*2*1

Well, by the Associative Law, we can group factors in any groupings, so we could insert parenthesis wherever we like. In particular , if I include a selection of factors that goes all the way to the right, all the way to 1, that’s another factorial. Thus:

8! = 8*(7*6*5*4*3*2*1) = 8*(7!)

8! = 8*7*(6*5*4*3*2*1) = 8*7*(6!)

8! = 8*7*6*(5*4*3*2*1) = 8*7*6*(5!) etc.

I chose 8! because we can see all the factors, but clearly we could extend this idea to any factorial, no matter how large:

237! = 237*236*(235!)

That’s an “unpacking” of the first two factors of 237! (BTW, this term, “unpacking”, is my own creation: you will not see this term used anywhere else.)

**Big Idea #3: when you divide two factorials, you “unpack” the larger one, and cancel it with the smaller one. **

Example:

**Big Idea #4: when you add/subtract two or more factorials, you unpack them all down to the lowest one, and factor out that common factor. **

For more on “factoring out”, see the section “Distributing and factor out” in this post.

Examples:

(200!) – (199!) = 200*(199!) – (1)*(199!) = (200 – 1)*(199!) = 199*(199!)

(200!) + (199!) = 200*(199!) + (1)*(199!) = (200 + 1)*(199!) = 201*(199!)

## Summary

Having read these rules, give those three practice problems another try before reading the solutions below. Here’s a fourth problem, with its own video explanation.

4) http://gmat.magoosh.com/questions/811

## Practice problem solutions

1) We are going to use the “factoring out” trick in both the numerator and the denominator:

Now, “unpack” that top factorial, to cancel the smaller one in the denominator:

Answer = **D**

2) “Unpack” the factorials in the numerator, so that everything is expressed as a product involving (89!) Then factor out that common factor.

Answer = **D**

3) Let’s consider the three expressions separately.

For expression I, we merely have to “unpack” the (49!) factor in the numerator.

So, expression I has a value of 49.

The expression II is tricky:

That’s a whole lot of factors in the numerator! That numerator is fantastically big: it has forty factors from 49 to 10 that are all greater than or equal to 10, so that means it’s automatically bigger than 10^40 (a number ** bigger than 1 trillion cubed** —

**OMG!**) That’s divided by 7! = 5040, so whatever this is, it’s way bigger than 49.

The expression III is a little easier than II:

That’s also a very big number. Each of the seven factors in the numerator is greater than 10, and 10^7 is ten million, so that numerator is more than 10,000,000. That’s divided by 7! = 5040. This is clearly bigger than 49. Notice, though, this has the same denominator as II, but many fewer factors in the numerator. Therefore II is much bigger than III.

Thus, from least to greatest, the order is I, III, II. Answer = **B**

Just as a note, if you are familiar with the idea of combinations, you may recognize expression II as a combinations number:

That would be the number of unique sets of 7 we could select from a pool of 49 unique items. There is no way you would be expected to calculate that number without a serious calculator or computer.

That’s a reasonably big number – just over 85 million. That’s a little more than a quarter the current population of the USA. For comparison, expression II is a real whopper:

This is a larger number than all humans and all other living things (animals & plants & all the way down to single-cell critters), including those alive now as well as those who have ever been alive on Earth. This number is larger than all the money in the world in pennies. This is more than the number of individual atoms comprising planet Earth and everything on Earth. This number is much much larger than the number of stars & planets & pulsars & quasars & black holes & whatever other star-like things in all galaxies & clusters in the visible Universe. The phrase “inconceivably big” does not even begin to capture how big this number is!

Hi Mike,

On your Big Idea #1, you stated “The number 73! must be divisible by 72!, by 47!, by 12!, etc.” Did you mean that The number 73! must be a multiple of 72!, 47!, or 12!? Or in other words, that 72!, 47!, or 12! must be divisible by 73!?

I don´t know if it was a typo or if I am misunderstanding the Number Properties wording (and Gmat has some creative ways to word these basics things in very tricky ways)

Thanks a lot!

Dear Wizzard,

I’m happy to respond. There’s no typo. When we say N is divisible by F, we are saying that N is a multiple of F, and F is a factor of N. For example, 35 is divisible by 5, so 35 is a multiple of 5, and 5 is a factor of 35. That’s always true. If the two numbers are not equal, the bigger one is the multiple, the smaller one is the factors, and the bigger one is divisible by the smaller one. Here, 73! is a bigger number than, say, 41!, and 73! is a multiple of 41!, which means that 41! is a factor of 73! It’s important to be crystal clear on the relationships of divisibility, multiples, and factors.

Does all this make sense?

Mike

Hi Mike,

You are absolutely correct. I apologize for my silly mistake. Since English is not my first language I sometimes get confused with the NP wording. I have to be extremely careful with this. As you were saying, this has to be crystal clear.

Apologies again.

Wizzard,

No problem, my friend. I’m glad we cleared things up. I wish you the very best of good fortune in your studies.

Mike

Mike,

On the second practice problem, I’m not understanding how you went from

91*90-90+1 to (91-1)*90+1

Can you help me?

Dear Cole,

In that step, I

factored outa factor of 90. The formal process isab – ac = a*(b – c)

When we move from left-to-right, that process is called “factoring out”, and when we move from right-to-left, it’s called “distributing” — these are the two sides of the Distributive Law, one of the huge fundamental laws of mathematics.

91*90 – 90 = (91)*(90) – (1)*(90) = (91 – 1)*90

Does all this make sense?

Mike

Yes it does now that I’ve looked at it again. I think the “+1” at the end was throwing me for a loop. Thanks!

Cole,

You are quite welcome. Best of luck to you.

Mike

For problem #3, I noticed the denominator in II was the product of perfect squares from 49 to 1 as follows:

denominator = (7!)^2 = (7 x 6 x 5 x 4 x 3 x 2 x 1) x (7 x 6 x 5 x 4 x 3 x 2 x 1)

re-arrange terms = (7 x 7) x (6 x 6) x (5 x 5) x (4 x 4) x (3 x 3) x (2 x 2) x (1 x 1), or

= (7^2) x (6^2) x (5^2) x (4^2) x (3^2) x (2^2) x (1^2)

denominator = 49 x 36 x 25 x 16 x 9 x 4 x 1

Substitution back into the original equation 49!/(denominator), I saw that I was only canceling the perfect squares in the factorial, so:

49!/(denominator) = 48 x 47 x 46 x … x 35 x 34 x … x 15 x … or all of the terms of 49! except perfect squares.

Therefor, it had to be far larger than I = 49 or III = the product of the first seven terms divided by 7!

Matt,

Yes, that’s a great way to think about it. Thanks for sharing it. Best of luck to you!

Mike

Hi Mike, I tried to solve the problem in the GMAT link that you gave. I have a doubt about it, I used your method and i got it right. But when i did the following i got the answer as 640, could you please explain why?

10! – 8! / 7!, this is what i did. 10* 9* (8!) – (8!) / 7!, now I cancel one 8! from 9* (8!), I get

10* 8* (8!) / 7!. How do i proceed from here? I did the following 10* 8 * 8 * 7! / 7!, then I cancelled 7! in the numerator and denominator and i get 640. What am i doing wrong?

Dear Amit,

First of all, you are completely ignoring grouping symbols, and this neglect leads to errors. See:

http://magoosh.com/gmat/2013/gmat-quant-mathematical-grouping-symbols/

Second, you ABSOLUTELY CANNOT CANCEL by subtraction a term within a product. If a term is part of a product, then it is 100% IMPOSSIBLE to cancel it with subtraction.

For simplicity, I would split it into two fractions:

(10! – 8!)/7! = (10!/7!) – (8!/7!)

= [(10*9*8*7!)/7!] – [(8*7!)/7!]

= 10*9*8 – 8 = 720 – 8 = 712

Does all this make sense?

Mike

“Thus, from greatest to least, the order is I, III, II. Answer = B”

Mike, in the conclusion of the 3rd problem you meant least to greatest, right?

-Preetam

Yes, I changed the typo. Thank you for pointing it out.

Mike

Hello Mike. Are these problems above the purview of the GRE exam, or are they likely to appear on the test?

Muhammad,

These are very hard problems, quite unlikely on either the GMAT or the GRE, but if you understand these, you will understand anything the GRE will throw at you regarding factorials.

Mike

Hello! Under the heading “Big Idea #2: you can “unpack” one factorial down to another.” it indicates:

***

I chose 8! because we can see all the factors, but clearly we could extend this idea to any factorial, no matter how large:

237! = 273*236*(235!)

***

I think you accidentally transposed “[…] 273*236 […]” and you intended to write “237! = 237*236*(235!)” — cheers!

Kris

Yes, you are 100% correct — that was a goofy mistyping on my part, and I just corrected it. Thank you very much.

Mike