How to calculate combination with 2 variables? Combination formula with 1 variable is simple $nCr = n! / (r!(n-r)!)$
So to find a number of combinations for rolling a 6 ONCE out of 6 rolls is
$$\dfrac {6! }{ (1!(6-1)!)} = 6$$
But how do you find number of combinations for rolling a 6 ONCE AND rolling a 4 TWICE out of 6 rolls?
I can't just do 6 choose 3 because they're different events
Thank you
 A: Assuming the die is fair, the probability of obtaining a 6, two fours, and three numbers other than 4 or 6 in that order is
$$\left(\frac{1}{6}\right)\left(\frac{1}{6}\right)^2\left(\frac{4}{6}\right)^3$$
However, we have to multiply that number by the number of orders in which one 6, two 4s, and three numbers other than 4 or 6 could occur in six rolls, which is
$$\binom{6}{1, 2, 3} = \binom{6}{1}\binom{5}{2}\binom{3}{3} = \frac{6!}{1!5!} \cdot \frac{5!}{2!3!} \cdot \frac{3!}{3!0!} = \frac{6!}{1!2!3!}$$
since there are six possible positions for the six, two of the remaining five positions must be filled with 4s, and all three of the remaining three positions must be filled by a number other than 4 or 6.  Hence, the desired probability is
$$\binom{6}{1, 2, 3}\left(\frac{1}{6}\right)\left(\frac{1}{6}\right)^2\left(\frac{4}{6}\right)^3$$
The number
$$\binom{6}{1, 2, 3}$$
is called a multinomial coefficient.  Since the probabilities are the same for each roll of the die, this is a multinomial distribution problem.
