A die roll problem

Suppose I roll a 6 sided die and a 4 sided die. How many different ways can the sum of my two die equal 7?

I want to say 4!/(1!(4-1)!). For each value on the 4 sided die, there is only one complementary value on the 6 sided die that will sum to 7.

I think 4 is correct but maybe the problem is easier than how you’ve written this.

if you want to write a fraction in LaTeX you use \frac {}{} the first brackets are the numerator and the second pair of brackets are the denominator, in this case, \frac {4!}{(1!(4-1)!)} \frac {4!}{(1!(4-1)!)} and for future reference, combinations are pretty much the same {4 \choose 1} {4 \choose 1}. In combinations, the brackets prevent anything you write near the combination from messing with it. This is without brackets (4 \choose 1) +2. (4 \choose 1) +2. This is with {4 \choose 1} +2 {4 \choose 1} +2

Would the answer here just be 4?

I think so but I’m trying to figure out what method that we’ve learned in class to use to get there.

I don’t think it’s a specific thing we learned in class. The simplest answer is 4 because there are only 4 ways to add the dice to get 7. I guess it could be the additive principle since the event of adding the dice can only be done in 4 ways but that seems like a stretch