Question

5vC3v

I really don't have any idea how to solve this equation. Look at the first post to see what is looks like on my paper.

Answer 1

\[_{5}C _{3}\]

Answer 2

\[ \large _{n}C_{k} = \frac{n!}{k! (n - k)!} \]

Answer 3

Which we use n = 5 and k = 3 in that expression to evaluate it

Answer 4

1.2?

Answer 5

No, these should come out to be integers.
If we write out the factorial:
\[ \frac{5\times4\times3\times2\times1}{(3\times2\times 1)(2\times1)}\]
There, we can just cancel out the things on the numerator and denominator. 3/3, 2/2, and 4/2 = 2/1

Answer 6

\[ \frac{5\times\cancel{4}2\times \cancel{3}\times\cancel{2}\times1}{(\cancel{3}\times\cancel{2}\times 1)(\cancel{2}\times1)} \]

Answer 7

so that just leaves 5*2 = 10

Answer 8

yeah, 10 thanks :D

Best answer, Closed. thanks again @AccessDenied !

Answer 9

You're welcome. :)