How do you simplify 6!4!5!8! ?

Question

Annie Levasseur · Accepted Answer

The factorial of any integer n is represented by

n \neq n (n - 1) (n - 2) \dots (3) (2) (1)

So, if we wish to calculate

\frac{6! 4!}{5! 8!}

the simplest way would be to expand each factorial, and cancel the common terms in the numerator and the denominator.

\frac{4! 6!}{5! 8!} = \frac{(4 \cdot 3 \cdot 2 \cdot 1) (6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1)}{(5 \cdot 4 \cdot 3 \cdot 2 \cdot 1) (8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1)}

If you look closely, all the terms in the numerator are cancelled out by terms in the denominator. Arranging them in this form also makes it easier to figure out which terms are getting cancelled. There are only three terms remaining in the denominator.

\frac{4! 6!}{5! 8!} = \frac{1}{(5) (8 \cdot 7)} = \frac{1}{280}

lovagwb · Accepted Answer

This is what I was looking for. Thanks!

How do you simplify \frac{6!4!}{5!8!} ?

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