Number and Algebra
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Calculus
Compute 8!6! 2!\frac{8!}{6!\,2!}6!2!8!
Simplify the expression 5!3! 2!\frac{5!}{3!\,2!}3!2!5!
Simplify 10!7! 3!\frac{10!}{7!\,3!}7!3!10!
Simplify 4!2!\frac{4!}{2!}2!4!
Express n!(n−2)! 2!\displaystyle\frac{n!}{(n-2)!\,2!}(n−2)!2!n! in simplest form
Simplify (n+2)!n! 2!\displaystyle\frac{(n+2)!}{n!\,2!}n!2!(n+2)!
Simplify 10!8!\frac{10!}{8!}8!10!
Evaluate 6!4! 2!+5!3! 2!\displaystyle\frac{6!}{4!\,2!}+\frac{5!}{3!\,2!}4!2!6!+3!2!5!
Compute 12!8! 4!\displaystyle\frac{12!}{8!\,4!}8!4!12!
Simplify (3!+2)!3!\frac{(3! + 2)!}{3!}3!(3!+2)!
Simplify (n+3)!(n−1)! 4!\displaystyle\frac{(n+3)!}{(n-1)!\,4!}(n−1)!4!(n+3)!
Simplify (n+1)!(n−1)! 3!\displaystyle\frac{(n+1)!}{(n-1)!\,3!}(n−1)!3!(n+1)!
Simplify 7!5! 4!\displaystyle\frac{7!}{5!\,4!}5!4!7!
For n>3n>3n>3, simplify n!/(n−3)!3!\displaystyle\frac{n!/(n-3)!}{3!}3!n!/(n−3)!
Simplify (n+4)!(n−2)! 6!\displaystyle\frac{(n+4)!}{(n-2)!\,6!}(n−2)!6!(n+4)!
Prove the identity (2n+1)!(n+1)! n!=2⋅(2n)!n! n!\displaystyle\frac{(2n+1)!}{(n+1)!\,n!}=2\cdot\frac{(2n)!}{n!\,n!}(n+1)!n!(2n+1)!=2⋅n!n!(2n)!
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Question Type 1: Calculating factorials
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Question Type 3: Working with expressions of factorials