Question

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

Answer 1

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

Explanation:

`"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 }`

`\qquad \qquad \qquad \qquad \qquad \qquad \qquad \ = \ { 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 ) }`

`\qquad \qquad \qquad \qquad \qquad \qquad \qquad \ = \ { 14 cdot ( 13 cdot 12 cdot 11 cdot cdots cdot 3 cdot 2 cdot 1 ) }/{ 1 cdot ( 13 cdot 12 cdot 11 cdot cdots cdot 3 cdot 2 cdot 1 ) }`

`\qquad \qquad \qquad \qquad \qquad \qquad \qquad \ = \ { 14 cdot color{red}cancel{ ( 13 cdot 12 cdot 11 cdot cdots cdot 3 cdot 2 cdot 1 ) } }/{ 1 cdot color{red}cancel{ ( 13 cdot 12 cdot 11 cdot cdots cdot 3 cdot 2 cdot 1 ) } }`

`\qquad \qquad \qquad \qquad \qquad \qquad \qquad \ = \ { 14 }/{ 1 }`

`\qquad \qquad \qquad \qquad \qquad \qquad \qquad \ = \ 14.`

`"Done !!"`

`"So we have our result:"`

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

Answer 2

`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

`(14xx13xx12xx11xx10xx...xx3xx2xx1)/(13xx12xx11xx10xx9xx...xx3xx2xx1)`

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!