A concept we have not yet covered in this series is factorials (though we used some factorials in the post Power in Factorials). Let’s first discuss the basics of factorials. Once we do, we will see that most factorial expressions can be easily solved using a single method: taking common!

First of all, what is (n!)?

n! = 1*2*3*4*5*6*…*(n – 2)*(n -1)*n

Let’s take some examples:

0! = 1 (mind you, it is not 0)

1! = 1

2! = 1*2

3! = 1*2*3

4! = 1*2*3*4

and so on…

Look carefully. Do you see any relation between 3! and 4!? Sure. 3! Appears in 4! too.

4! = 1*2*3*4 = (1*2*3)*4 = (3!)*4

Similarly, 2! is also a part of 3! as well as 4!

4! = 1*2*3*4 = (2!)*3*4 = (3!)*4

As a general note, we can say that:

n! = (n – 1)! * n

n! = (n – 2)! * (n – 1) * n

n! = (n – 3)! * (n – 2) * (n – 1) * n

and so on…

We can write n! in many different ways. We use whatever suits us best in the question. How does knowing this help us solve questions? Let’s see:

Question: If (n-2)!= [n! + (n-1)!]/99 and n is a positive integer, how many distinct values can n take?

(A) 0
(B) 1
(C) 2
(D) 3
(E) Infinite

Solution: We need to solve this equation to find out the values of x which satisfy it. But how do we solve equations with factorials in them?

(n-2)!= [n! + (n-1)!]/99

It looks rather complicated, right? It is not, actually! Let’s use what we have just learned. We need to separate the factorials from the rest of the equation. To do that, we need to take something common. The left hand side has (n-2)!

We know that n! = (n-2)!*(n-1)*n

and (n-1)! = (n-2)!*(n-1)

The equation becomes: (n-2)!= [(n-2)!*(n-1)*n + (n-2)!*(n-1)]/99

(n-2)! = (n-2)!*[(n-1)*n + (n-1)]/99

99(n-2)! = (n-2)!*[(n-1)*n + (n-1)]

(n-2)!*[(n-1)*n + (n-1) – 99] = 0

The product of two factors (n-2)! and [(n-1)*n + (n-1) – 99] must be 0 so at least one of them must be 0. Notice that factorial of a number cannot be 0 so the other factor i.e. [(n-1)*n + (n-1) – 99] must be 0.

[(n-1)*n + (n-1) – 99] = 0

n^2 = 100

n can take two values: 10 and – 10

But it is given to us that n is a positive integer so only 10 is acceptable.

Hence, there is only 1 value which satisfies this equation.

Answer (B)

Remember, when dealing with multiple factorials, all you can do is take something common. But then, that may be all you need to do!

Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep in Detroit, and regularly participates in content development projects such as this blog

### 5 Responses

1. Amatuer says:

when you got n=10, -10…. isn’t -10 automatically thrown away because factorials are not defined for negative numbers!? why do we even need to look at the question for him to mention that n is positive integer?

• Karishma says:

You can use either logic to reject n = -10

2. Anthony says:

Hello Karishma

– I understand that [(n-1)*n + (n-1) – 99] must be zero since the factorial (n-2)! cannot be equal to zero, but, why is n^2 = 100? And from where do come from this n^2)

– And from where do we know that (n-1)! = (n-2)!*(n-1) ? (is it part of a previous operation on the n! = (n-2)!*(n-1)*n?)?

Thank you

• Karishma says:

(n-1)*n + (n-1) – 99 = 0
n^2 – n + n -1 – 99 = 0 (simplifying the equation)
n^2 – 100 = 0
n^2 = 100

Do you understand that n! = (n – 1)! * n ?
i.e. 5! = 4!*5

i.e. factorial of a number is equal to the product of the factorial of the previous number and that number (except the case where the number is 0 because 0! = 1)
Similarly, factorial of (n-1) will be product of (n-2)! and (n-1)

• Anthony says:

Ok, I totally get it now, I wrote it down on a paper and everything went clear (with your explanations).

Thank you Karishma