# Math Help - Factorial Form

1. ## Factorial Form

Using the factorial form of the binomial theorem, how can you calculate the answer when the numbers just get too big for the calculator? For example, I'm trying to do 30! / 28! 2!

Is there a simpler way of trying to work out these monstrous problems?

2. You have to simplify the factorials.
$\displaystyle \frac{30!}{28!2!}=\frac{28!\cdot 29\cdot 30}{28!2!}=\frac{29\cdot 30}{2}=29\cdot 15=435$

3. Hello, BlueStar!

Using the factorial form of the binomial theorem,
how can you calculate the answer when the numbers just get too big for the calculator?
For example, I'm trying to do: $\frac{30!}{28!2!}$

Are you new to factorials?
If you're that new, you shouldn't be working with binomial coefficients.
First, you need practice on some basics.

If you were given: . $\frac{7!}{6!}$, would you really mutliply it all out?

. . $\frac{8!}{7!} \;=\;\frac{8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\c dot1}{7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1} \;\begin{array}{cc}\Rightarrow \\ \Rightarrow\end{array} \;\frac{40,320}{5040}$

Then on your calculator: . $40,320 \div 5040 \;=\;8$ ?

Or would you notice that there is a lot of possible cancelling:

. . $\frac{8\cdot\not7\cdot\not6\cdot\not5\cdot\not4\cd ot\not3\cdot\not2\cdot\not1}{\not7\cdot\not6\cdot\ not5\cdot\not4\cdot\not3\cdot\not2\cdot\not1} \;=\;8$

Back to your problem . . . $\frac{30!}{28!2!}$

Look at what we have: . $\frac{30\cdot29\cdot28\cdot27\cdot26\cdots2\cdot1} {(28\cdot27\cdot26\cdots2\cdot1)(2\cdot1)}$

Reduce: . $\frac{30\cdot29\,\cdot\not2\!\!\!\not8\,\cdot\not2 \!\!\!\not7\,\cdot\not2\!\!\!\not6\cdots\not2\cdot \not1}{(\not2\!\!\!\not8\,\cdot\not2\!\!\!\not7\,\ cdot\not2\!\!\!\not6\cdots\not2\cdot\not1)(2\cdot1 )} \;=\;\frac{30\cdot29}{2\cdot1} \;=\;15\cdot29 \;=\;435$

See? . . . Look ahead and we don't have to write out all the numbers.

The number of five=card poker hands is: . $\frac{52!}{5!47!}$

We start writing the numerator: . $52\cdot51\cdot50\cdot49\cdot48\cdots$
. . but we know that the factors from 47-down-to-1 will cancel out.

All we have left is: . $\frac{52\cdot51\cdot50\cdot49\cdot48}{5\cdot4\cdot 3\cdot2\cdot1}$

. . which reduces further: . $\frac{\not5\!\!\!\not2^{13}\,\cdot\not5\!\!\!\not1 ^{17}\,\cdot\not5\!\!\!\not0^{10}\cdot49\,\cdot\no t4\!\!\!\not8^{24}}{\not5\,\cdot\not4\,\cdot\not3\ ,\cdot\not2\cdot1} \;=\;2,598,960$