Simplify, please: \frac{n!}{(n-3)!}

Alfred Martin 2021-12-20 Answered
Simplify, please:
\(\displaystyle{\frac{{{n}!}}{{{\left({n}-{3}\right)}!}}}\)

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

Jim Hunt
Answered 2021-12-21 Author has 2151 answers
\(\displaystyle{\frac{{{n}!}}{{{\left({n}-{3}\right)}!}}}\)
\(\displaystyle={\frac{{{n}{\left({n}-{1}\right)}{\left({n}-{2}\right)}{\left({n}-{3}\right)}{\left({n}-{4}\right)}\ldots.\times{3}\times{2}\times{1}}}{{{\left({n}-{3}\right)}{\left({n}-{4}\right)}\ldots\times{3}\times{2}\times{1}}}}\)
\(\displaystyle={n}{\left({n}-{1}\right)}{\left({n}-{2}\right)}\)
\(\displaystyle{n}{\left({n}^{{{2}}}-{3}{n}+{2}\right)}\)
\(\displaystyle={n}^{{{2}}}-{3}{n}^{{{2}}}+{2}{n}\)
Not exactly what you’re looking for?
Ask My Question
0
 
Paineow
Answered 2021-12-22 Author has 1356 answers
\(\displaystyle{\frac{{{n}!}}{{{\left({n}-{3}\right)}!}}}\)
Cancel the factorials: \(\displaystyle{\frac{{{n}!}}{{{\left({n}-{m}\right)}!}}}={n}\times{\left({n}-{1}\right)}\ldots{\left({n}-{m}+{1}\right)}\), n>m
Thus, we have
\(\displaystyle{\frac{{{n}!}}{{{\left({n}-{3}\right)}!}}}={n}{\left({n}-{1}\right)}{\left({n}-{2}\right)}\)
0
RizerMix
Answered 2021-12-29 Author has 9821 answers

\(\frac{n!}{(n-3)!}=n(n-1)(n-2)=n^{2}-3n^{2}+2n\)

0

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question
...