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

2 Answers
Mar 6, 2018

"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.

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

(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!