To solve it, simply remember that a factorial (some value such as 9 with an exclamation mark afterwards) means that you would take it to be (9x8x7x6x5x4x3x2x1). Since in your question you have another value in the denominator that is a factorial smaller than 9! some of the terms will cancel out making the problem more manageable if done by hand.
For example 9!/5! = (9x8x7x6x5x4x3x2x1)/(5x4x3x2x1) which simplifies to (9x8x7x6) since the other values cancel.
By the way, the quickest way to evaluate something like 6!9! is to recognize that 9!= 9(8)(7)(6)(5)(4)(3)(2)(1)= 9(8)(7)(6!) so that 6!9!=9(8)(7). Now to find 6!3!9! write it as (3)(2)9(8)(7) and cancel.
(This is a "binomial coefficient", i!(n−i)!n!, also written "NCi" or (ni), with n= 9 and i either 6 or 3.)
(Lookagain is completely correct as to the standard interpretation of what you wrote. I just immediately jumped to what would be a more common problem.)
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