# How do you simplify (14!)/(13!)?

Mar 6, 2018

 "This is the result:" \qquad \qquad \qquad \quad { 14! }/{ 13! } \ = \ 14.

#### Explanation:

$\text{Watch how this goes -- these can be fun ... }$

 \qquad \qquad \qquad \qquad \qquad \ { 14! }/{ 13! } \ = \ { 14 cdot 13 cdot 12 cdot 11 cdot cdots cdot 3 cdot 2 cdot 1 }/{ 13 cdot 12 cdot 11 cdot cdots cdot 3 cdot 2 cdot 1 }

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus = \setminus \frac{14 \cdot \left(13 \cdot 12 \cdot 11 \cdot \cdots \cdot 3 \cdot 2 \cdot 1\right)}{\left(13 \cdot 12 \cdot 11 \cdot \cdots \cdot 3 \cdot 2 \cdot 1\right)}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus = \setminus \frac{14 \cdot \left(13 \cdot 12 \cdot 11 \cdot \cdots \cdot 3 \cdot 2 \cdot 1\right)}{1 \cdot \left(13 \cdot 12 \cdot 11 \cdot \cdots \cdot 3 \cdot 2 \cdot 1\right)}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus = \setminus \frac{14 \cdot \textcolor{red}{\cancel{\left(13 \cdot 12 \cdot 11 \cdot \cdots \cdot 3 \cdot 2 \cdot 1\right)}}}{1 \cdot \textcolor{red}{\cancel{\left(13 \cdot 12 \cdot 11 \cdot \cdots \cdot 3 \cdot 2 \cdot 1\right)}}}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus = \setminus \frac{14}{1}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus = \setminus 14.$

$\text{Done !!}$

$\text{So we have our result:}$

 \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ { 14! }/{ 13! } \ = \ 14.

Jul 8, 2018

$14$

#### Explanation:

The answer to this question below is a more systematic approach that's great, especially if you're new to factorials, but we have a special case here:

(14!)/(13!)

This is in the form

((a+1)!)/(a!), where in our example, $a = 13$.

If we were to expand this out, every term would cancel except for the first one. Here's what I mean:

In our example, we essentially have

$\frac{14 \times 13 \times 12 \times 11 \times 10 \times \ldots \times 3 \times 2 \times 1}{13 \times 12 \times 11 \times 10 \times 9 \times \ldots \times 3 \times 2 \times 1}$

Notice, every term on the top and bottom would cancel except for the $14$, because every term the denominator has, so does the numerator.

In general, ((a+1)!)/(a!) simplifies to $a$, so if you have

(21!)/(20!), this would be $21$.

If we had (47!)/(46!), this would be $47$.

Hope this helps!