Because flipping a coin and looking for heads is a clear example of a success or failure situation, and the number of tosses is fixed, X is a binomial random variable. (It's binomial because each trial, each attempt to flip the coin, has two outcomes).
Since you toss the coin 5 times, the most number of heads you could get is 5. The fewest you could get is 0.
There's a recurring concept in probability that comes up here: Combinations and Permutations. The notation we use when we don't have fancy layout available is this:
nCr means "The number of ways you can pick r things from a group of n things when order doesn't matter"
nPr means "The number of ways you can pick r things from a group of n things when order matters"
Let's say I have 5 objects, labeled 1,2,3,4,5, and I want to pick 2 of them.
5C2 would be the number of ways I could do this. I have 5 choices for my first number, and 4 for my second number. But I could pick 1 then 2 or 2 then 1, and I'd have counted all the possibilities twice. I have to divide by two to get my final answer: 5x4/2 = 10.
5P2 is the number of ways I could pick them if order did matter: 1 then 2 would be different from 2 then 1.
5P2 = 5x4 = 20.
Going back to your coin flip problem, the number of ways I can get 2 heads out of the 5 coins is 5C2. The number of ways I can get 3 heads out of the 5 coins is 5C3.
So your distribution is a discrete random variable X where X can take on the value from 0 through 5 with the probability of any particular value x being: 5Cx (1/2)^5
P(X=0)= 5C0 (1/2)^5 = 1/32.
P(X=1)= 5C1 (1/2)^5 = 5/32.
P(X=2)= 5C2 (1/2)^5 = 10/32.
P(X=3)= 5C3 (1/2)^5 = 10/32.
P(X=4)= 5C4 (1/2)^5 = 5/32.
P(X=5)= 5C5 (1/2)^5 = 5/32.
Notice the the probability of X being a value between 0 and 5 inclusive is 32/32 = 1. This should always happen.
