How do you simplify the factorial expression #(4!)/(6!)#?

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Answer 1 · 2017-03-08T09:28:58

#1/(6xx5)=1/30#

We can expand factorials into their constituent multiplication series:

#(4xx3xx2xx1)/(6xx5xx4xx3xx2xx1)#

We can then cancel out the #4, 3, 2, and 1# in both denominator and numerator and we're left with

#1/(6xx5)=1/30#

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We could also express our initial problem as:

#(4!)/(6xx5xx4!)#, cancel out the #4!# terms, and be left with #1/30#