Find the total number of elements that satisfy this condition..

eddy2017 said:
I'm sorry. This is the first time I do this type of exercise.
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[math]\frac{\text{n!}}{3!(n-3)!}=\frac{n(n-1)(n-2)\sout{(n-3)(n-4)(n-5)...(3)(2)(1)}}{6\sout{(n-3)(n-4)(n-5)...(3)(2)(1)}}[/math]I was hoping you would notice and continue the factorial expansion in the numerator and see most of the terms cancel out.
What you have left is [math]\frac{n(n-1)(n-2)}{6}=20[/math]This is how pka derived the equation in the previous post.
 
Thank you so much for explaining patiently. As I told you my experience with factorial has only been with permutation and combination operations
 
I will study about how to find factorial of expressions with variables in it. First thing tomorrow.
 
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