What math technique was used to proceed from step 1 to step 2?

Question

In my Math textbook under the Permutations unit, one of the example questions is as follows: Show that 100! + 99! = 101(99!) without using technology.

After the question, the example breaks the question down for you as done below:

100! + 99! = 100(99!) + 99!

= 99!(100+1)

= 99!(101)

= 101(99!)

Now, my source of confusion stems from the transition from step 1 to step 2. Step 1 being the first line of the solution and step 2 being the second. I understand 100! is the same as 100(99!), but everything else after that doesn't make any sense to me. Could someone help clarify this for me?

Thank you.

Answer 1

\begin{align} 100! + 99! &= 100(99!) + 99! \tag1\\ &= 99!(100+1) \tag2\\ &= 99!(101) \tag3\\ &= 101(99!) \tag4 \end{align}

• $(2)$ is distributive law $ab + ac = a (b + c)$ with $a = 99!$, $b = 100$ and $c = 1$.
• $(4)$ is because multiplication is commutative: $ab = ba$

Answer 2

To go to the second line they used the fact that $99!=1(99!)$ and the distributive principle. There is a common factor of $99!$ in both terms on the right of line 1. From line 2 to line 3 they just added and to line 4 they used the commutative principle. If they hadn't commuted the terms in going from line 1 to line 2 they wouldn't have had the third line.