How do you evaluate #(14!)/(5!9!)#?

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Answer 1 · 2016-10-28T12:51:47

#(14!)/(9!xx5!)=2002#

#n! = nxx(n-1)xx(n-2)xx(n-3)xx...........xx3xx2xx1#

Hence #14! = 14xx13xx12xx11xx...3xx2xx1#

#9! = 9xx8xx7xx6xx5xx4xx3xx2xx1# and #5!= 5xx4xx3xx2xx1#

Hence #(14!)/(9!xx5!)=(14xx13xx12xx11xx10xx9xx8xx7xx6xx5xx4xx3xx2xx1)/(9xx8xx7xx6xx5xx4xx3xx2xx1xx5xx4xx3xx2xx1)#

= #(14xx13xx12xx11xx10xxcancel(9xx8xx7xx6xx5xx4xx3xx2xx1))/(cancel(9xx8xx7xx6xx5xx4xx3xx2xx1)xx5xx4xx3xx2xx1)#

= #(14xx13xx12xx11xx10)/(5xx4xx3xx2xx1)#

= #(14xx13xxcancel12xx11xxcancel5cancel10)/(cancel5xxcancel(4xx3)xxcancel2xx1)#

= #14xx13xx11#

= #2002#