Question

What is 6c3 simplified?

Answer 1

Combinations?

Answer 2

6c3 written in different form is:
\[\left(\begin{matrix}6 \\ 3\end{matrix}\right)\]
and formula for that is:
\[\left(\begin{matrix}n \\ k\end{matrix}\right) = \frac{ n! }{ k!*(n-k)! }\]

Answer 3

now you know 6 is n and 3 is k. plug the numbers into the formula

Answer 4

am going to make an assumption that "6c3" is referring to "Combinations" notation, ₆C₂ ,

which is the answer to the Q:

"How many ways can we choose 2 distinct objects from 6 distinct objects, where order is NOT important?"

For example, given the letters A, B, C, D, E, and F, How many different two-letter combinations can be we make? (but where AB is equivalent to BA, so don't count BA as being different from AB)

₆C₂ = 6! ÷ [(6–2)! x 2!]

₆C₂ = 6! ÷ [4! x 2!]

₆C₂ = 720 ÷ [24 x 2]

₆C₂ = 720 ÷ 48

₆C₂ = 15

➞ There are 15 ways can we choose 2 distinct objects from 6 distinct objects, where order is NOT important.

Answer 5

Ask yahoo? Knowel

Answer 6

no yahoo sucks

Answer 7

my uncle taught me that example