How do you simplify #(19!)/(13!)#?

1 Answer
Oct 1, 2016

#19,530,040#

Explanation:

#!# reads as "factorial" It means multiply as shown in the example:

#6! = 6xx5xx4xx3xx2xx1 = 720#

Because the numbers become so very big, you ideally need a scientific calculator.

However, there is some simplifying you can do ..
#(19!)/(13!) = (19xx18xx.......xx14xx(color(red)(13 xx ..... xx1)))/color(red)((13xx12xx11...xx1))#

#13!# forms part of #19!# as shown above, so these factors cancel....

#(19!)/(13!) = (19xx18xx.......xx14xxcancel(color(red)(13 xx ..... xx1)))/cancelcolor(red)((13xx12xx11...xx1))#

leaving only #19xx18xx17xx16xx15xx14# to calculate, which gives

#19,530,040" "larr# still a horribly big number!