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

1 Answer
mason m
Sep 7, 2016

56

Explanation:

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

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