# How does factorials work in algebraic equations?

I was reading a text book and found these two lines and I have no clue how did step1 become step2. Please help me with this. Thanks

Step 1:

$$\frac {5!}{(4-r)!} = \frac {6*5!}{(5-r+1)(5-r)(5-r-1)!}$$

Step 2: $$(6-r)(5-r)=6$$

Text book: (NCERT 11th Chapter 7, Example 13)

If we simplify the first step, we have $$\frac{5!}{(4-r)!}=\frac{6 \times 5!}{(6-r)(5-r)(4-r)!}$$
Now multiply each side by $\frac{(6-r)(5-4)(4-r)!}{5!}$.
Thus we have $$\frac {5!}{(4-r)!} = \frac {6*5!}{(5-r+1)(5-r)(5-r-1)!} \Leftrightarrow (6-r)(5-r)=6$$
• @PYmacstue It makes it easier to simplify. The right side has $(4-r)!$, so we have to multiply each side by $(4-r)!$. But if the right side if $(6-r)!$, It's harder to see. – S.C.B. Mar 28 '16 at 9:06