Question

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

Answer 1

\[n!=n\times(n-1)!\]

Answer 2

what is n

Answer 3

some natural number

Answer 4

so if its not a number how would i solve this

Answer 5

Well you're diverting yourself to your own question here.

Answer 6

\[\frac{\color{teal}{7!}\times8!\times3!}{5!\times\color{orange}{9!}}=\frac{\color{teal}{7\times6\times5!}\times8!\times3!}{5!\times\color{orange}{9\times8!}}\]

Answer 7

72576

Answer 8

\[n!=n\times(n-1)!\]\[\quad=n\times(n-1)\times(n-2)!\]\[\quad=n\times(n-1)\times(n-2)\times\cdots\times3\times2\times1\]

Answer 9

no still not getting

Answer 10

for example
\[4!=4\times3\times2\times1\]\[6!=6\times5\times4\times3\times2\times1\]


\[\frac{4!}{6!}=\frac{4\times3\times2\times1}{6\times5\times4\times3\times2\times1}=\frac{4!}{6\times5\times4!}=\frac1{6\times5}=\frac1{30}\]

Answer 11

the trick is to cancel common factors ,

Answer 12

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

Answer 13

\[\frac{{7!}\times8!\times3!}{5!\times{9!}}=\frac{{7\times6\times5!}\times8!\times3!}{5!\times{9\times8!}}=\frac{{7\times6\times\cancel{5!}}\times\cancel{8!}\times3!}{\cancel{5!}\times9\times\cancel{8!}}\]

Answer 14

so 14

Answer 15

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

Answer 16

i choose not to fully expand the factorials because it takes to much space and time,

bytheway 14 isn't quite right

Answer 17

what its not that was one of the options

Answer 18

can you show me your working

Answer 19

@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.