# Why is the permutations of 6 objects where 2 are similar and the other 4 are similar is the same as 6c2?

I'm studying binomial expansion, and to get all the outcomes or arrangements of head appearing twice after throwing a coin 6 times without using a tree diagram, it's written that we can use 6c2 to get the answer.

But I totally don't understand how 6!/2!4! is the same thing as 6c2. I need order to matter, here, I don't want for instance THHTTTT to be the same as THTTTHT. And when I read 6c2 the only thing I can imagine is two empty spaces (_ _) and me trying to make combinations of 6 objects in them.

• ${7 \choose 3} = \frac{7!}{3!(7-3)!} = \frac{7!}{3!\cdot 4!}$ by definition. – BruceET Sep 26 at 5:36