Q&A for How to Multiply Factorials

There is no general rule covering this situation. In your example, however, (3!)(5!) = 720 = 6!.

Multiplying by 2! means multiplying by 2.

Multiply (k+1)! by two.

Do it on a computer with a high precision integer arithmetic program if you need the exact number written out in full. It took a few milliseconds of CPU time on mine. 1000! is a 2568 digit number beginning with 402,387,260,..., while 500! is a 1135 digit number beginning with 1,220,136,...

Factorial equations are hard to simplify, and positive integer solutions to a!b! = c!d! are rare enough that your equations is more likely to have a trivial solution (one where {a,b} = {c,d}) than a tricky one. Let's start there and see if we get lucky. We can't have (8-r) = (10-r), but (8-r) = r holds when r = 4. And if r = 4, then (r+2) = (10 - r) = 6. So r=4 is a solution because 4!6! = 6!4!. An alternative solution which might be a slicker way to prove the above solution is unique is to divide both sides by 10! and look at the combinatorial interpretation. If you flip 10 fair coins the left side is the probability of (8-r) heads and (r+2) tails while the right side is the probability of.

n x n! = n x n! + n! - n! = n x n! + 1 x n! - n! = (n+1) x n! - n! = (n+1)! - n!. The first few steps of that derivation are just basic algebraic manipulations: add 0 (in the form n!-n!) and the distributive law. The last is the recursive definition of factorial (6! = 6 x 5!).

13! + 14! = 2(13!) + 14.

That equals (2)(5)(4)(3)(2)(1) = 240.

First evaluate the factorial. Then multiply that product by the constant. For example, (4)(6!) = (4)[(6)(5)(4)(3)(2)(1)] = (4)(72) = 2,880.

Think about it like this: n!(n+1) is just (n+1)(n)(n-1)(n-2)...(1), which is also (n+1)!. Write out a few terms of n! and move the (n+1) term to the front so that it's easier to see that n!(n+1) is the same as (n+1)!.

(99!)(99!) = (9.33 x 10^155)² = 8.7 x 10^311 (approximately).