Help me evaluate this question: Evaluate 9!/3!6!

[SaraLove]

Help me evaluate this question

Evaluate 9!/3!6!

 

[ZibZabZabbityDoo]

that's not a calculus problem.

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.

 

[HallsofIvy]

By the way, the quickest way to evaluate something like \(\displaystyle \frac{9!}{6!}\) is to recognize that 9!= 9(8)(7)(6)(5)(4)(3)(2)(1)= 9(8)(7)(6!) so that \(\displaystyle \frac{9!}{6!}= 9(8)(7)\). Now to find \(\displaystyle \frac{9!}{6!3!}\) write it as \(\displaystyle \frac{9(8)(7)}{(3)(2)}\) and cancel.

(This is a "binomial coefficient", \(\displaystyle \frac{n!}{i!(n-i)!}\), also written "\(\displaystyle _NC_i\)" or \(\displaystyle \begin{pmatrix}n \\ i\end{pmatrix}\), 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.)

 

[lookagain]

Evaluate 9!/3!6!

SaraLove,

what I'm sure you intended as the problem, you typed incorrectly.

What you actually typed is equal to \(\displaystyle \bigg(\dfrac{9!}{3!}\bigg)6! \ = \ 43,545,600.\)


You needed to have used grouping symbols, such as in the following:

9!/(3!6!)




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