Factorial Problems

Question

Please explain + show steps!! (mainly confused about a,b,c)

a) (n+2)(n+1)!

b) (n+1)! / (n-1)!

c) n(n-1)!

d) 2 x 5! / 3! 2! (The 2 is not with the 5!, its the multiplying the whole equation)

Answer 1

The trick of doing this is writing out the factorials. n! can be written as n*(n-1)*(n-2)*...* 3*2*1. You start with n and then decrease your number by 1 each time until you get to 1.

a) (n+2) * (n+1)! = (n+2) * (n+1) * n *...* 3 * 2 * 1. This is the same as (n+2)!

b) (n+1)! / (n- 1)! = [(n+1) * n * (n-1)* (n-2) * ...* 3 * 2 * 1] / [ (n-1)* (n-2) * ...* 3 * 2 * 1]

Notice that a lot of this cancels. There's a 1 in both the numerator and the denominator so they cancel. The 2's cancel and the 3's. Every number up to and including n-1 will cancel, so you'll be left with just (n+1)*n

c) This one is similar to part a).

Answer 2

It looks like the questions are asking you to simplify each expression. The key to each of these is knowing the general form of a factorial:

n! = n(n-1)(n-2)(n-3). . .(3)(2)(1)

or

9! = 9x8x7x6x5x4x3x2x1

a) (n+2)(n+1)!

If we expand out the (n+1)!,

(n+2)(n+1)(n)(n-1)(n-2). . .(3)(2)(1)

Notice that this is the same as (n+2)!, therefore,

(n+2)(n+1)! = (n+2)!

b) (n+1)! / (n-1)!

This problem focuses more on cancelling between the numerator and denominator, let's expend them out first.

(n+1)! / (n-1)! = (n+1)(n)(n-1)(n-2). . .(3)(2)(1) / (n-1)(n-2). . .(3)(2)(1)

Notice that the numbers (n-1)(n-2). . .(3)(2)(1) appears in both the numerator and denominator, so we can cancel them out.

(n+1)! / (n-1)! = (n+1)(n) = n² + n

c) n(n-1)!

This problem is almost identical to part a, expand out and notice that it is exactly the same as n!