Factorial to Permutation; (n+3)! = 720*n!

Question

The problem is : Find $n$ if:

$$(n+3)! = 720 \cdot n!$$

The explanation goes as following:

a) $\frac{(n+3)!}{n!}=720$

b) $_{(n+3)}\text{P}_3 = 720$

c) $_{10}P_3 = 720$

d) $n=7$

What I don't get is how it went from step (a) to (b) and where the $3$ come from?

Answer 1

a) $$\frac{(n+3)!}{n!}=720$$ By definition of factorials $$\frac{(n+3)(n+2)(n+1)[(n)!])}{n!}=720$$ Dividing the numerator and denominator by n! $$(n+3)(n+2)(n+1)=720$$ By definition of $nPk=\frac{n!}{(n-k)!}$ $$(n+3)P3 = \frac{(n+3)!}{(n+3-3)!}=\frac{(n+3)(n+2)(n+1)[(n)!]}{(n)!}=(n+3)(n+2)(n+1)$$ So using $(n+3)P3$ b) $$(n+3)P3 = 720$$

Answer 2

From your comment, it looks like you know what $nPk$ means, so:

$$\frac{(n+3)!}{n!}=\frac{(n+3)!}{[(n+3)-3]!}=_{n+3}P_{3}$$

Answer 3

$$(n+3)! = 720 \cdot n! \implies \frac {(n+3)!}{n!}=720$$

$$(n+3)(n+2)(n+1)=720 \implies (n+3)P3 =720 $$

$$ 720=(10)(9)(8) \implies n=7$$