How do you simplify #(7!)/(2!5!)+(7!)/(4!3!)#?

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Answer 1 · 2016-09-07T11:08:18

#56#

Since #n! =n(n-1)!#, that is to say, the product of #n# and all the integers below it, until #1#, first make sure you understand this simplification:

#(10!)/(8!)=(10*9*8!)/(8!)=10*9#

Both #10!# and #8!# have the #8!# chain within them, just #10!# additionally has the #9# and #10#.

Going back to the original expression:

#(7!)/(2!5!)+(7!)/(4!3!)#

Cancel the #5!# with the #7!#, leaving #7*6#, and in the other fraction cancel the #4!# and the #7!#, leaving #7*6*5#:

#=(7*6)/(2!)+(7*6*5)/(3!)#

Write out #2!# and #3!#:

#=(7*6)/(2*1)+(7*6*5)/(3*2*1)#

Notice that #3! =6#, so it cancels straightaway with the #6# in the numerator. In the first fraction, just take divide #2# from the #6#.

#=7*3+7*5#

#=56#