How do you simplify the factorial expression #(10!)/(8!)#?

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Answer 1 · 2017-02-07T06:55:22

#90.#

Recall that, #(n!)=n(n-1)(n-2)...3xx2xx1,#

#=n{(n-1)(n-2)(n-3)...3xx2xx1}#

#rArr (n!)=n{(n-1)!}...............(ast)#

#:. (10!)/(8!)={10(9!)}/(8!), ...............[because, (ast)]#

#=[(10){9(cancel(8!))}]/cancel(8!)#

#=(10)(9)#

#=90.#