n?k denotes the product of all integers from n to k, inclusive.

Let's take a small number to understand the concept

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

= 1 * 2 * 3 * (2*2) * 5 * (3*2)

The power of a number in a factorial is the number of times that number appears in the factorial.

For example - In 6! the power of 3 is 2, because 3 appears two times, indicated in red, in 6!

Put in other words, we can say that 6! is divisible by \(3^2\).

In a similar manner the power of 2 in 6! is 4. This is again because of the same reason, that 2 appears 4 times in 6! (indicated in blue)

In other words 6! is divisible by \(2^4\).

Let's take another number, say 9!

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

= 1 * 2 * 3 * (2*2) * 5 * (2*3) * 7 * (2*2*2) * (3*3)

Power of 2 in 9! = 7
Power of 3 in 9! = 4

Hope this is clear !

Ok I understand it more now. How were you able to figure out that the power of 3 in 103 is 49 so quickly? What gave it away?

Please refer to this post by Bunuel

You may want to read "Finding the number of powers of a prime number p, in the n!" under Factorial section of the post.

Ok I finally understand the concept! Thanks for posting that link gmatophobia. One last quick question why was 91 used in the orginal explanation and not 92? Since the question states 92...

I am assuming you're referring to the 91! that's in the denominator.



We have divided 103! by 91! as the product starts from 92.

Had we divided 103! by 92!, the 92 in the numerator would have got cancelled out and the product would start from 93.

To understand this we can take the below example -

Ex. Let's say the product is 6 * 7 * 8 * 9

We can write the expression as \(\frac{9! }{ 5!} \) = \(\frac{1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 }{ 1 * 2 * 3 * 4 * 5 }\)

The common terms get cancelled out and we're left with 6 * 7 * 8 * 9

On the contrary, if we divided by 6!, the product that we will end up with is 7 * 8 * 9 (and not 6 * 7 * 8 * 9).

Hope this clarifies.