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7!*8!*3!/5!*9!

Mathematics
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\[n!=n\times(n-1)!\]
what is n
some natural number

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Other answers:

so if its not a number how would i solve this
Well you're diverting yourself to your own question here.
\[\frac{\color{teal}{7!}\times8!\times3!}{5!\times\color{orange}{9!}}=\frac{\color{teal}{7\times6\times5!}\times8!\times3!}{5!\times\color{orange}{9\times8!}}\]
72576
\[n!=n\times(n-1)!\]\[\quad=n\times(n-1)\times(n-2)!\]\[\quad=n\times(n-1)\times(n-2)\times\cdots\times3\times2\times1\]
no still not getting
for example \[4!=4\times3\times2\times1\]\[6!=6\times5\times4\times3\times2\times1\] \[\frac{4!}{6!}=\frac{4\times3\times2\times1}{6\times5\times4\times3\times2\times1}=\frac{4!}{6\times5\times4!}=\frac1{6\times5}=\frac1{30}\]
the trick is to cancel common factors ,
i subtract the 7 and 3 on the top so i would have 8*6*5*4*2*1/8*7*6*4*3*2*1
\[\frac{{7!}\times8!\times3!}{5!\times{9!}}=\frac{{7\times6\times5!}\times8!\times3!}{5!\times{9\times8!}}=\frac{{7\times6\times\cancel{5!}}\times\cancel{8!}\times3!}{\cancel{5!}\times9\times\cancel{8!}}\]
so 14
but why didnt you right it out all the way as in 7*6*5*4*3*2*1 for all the numbers on top and bottom then facter them out
i choose not to fully expand the factorials because it takes to much space and time, bytheway 14 isn't quite right
what its not that was one of the options
can you show me your working
@eugeniaday if you do these on your calculator, make sure you use parentheses! 7!*8!*3!/5!*9! is NOT the same as 7!*8!*3!/(5!*9!), and your calculator can't instinctively know what you *meant* to enter.

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