The $50:50$ odds on each toss means that all distinct outcomes are equally weighted, and so we don't worry about it any further. Here an outcome is an arrangement of heads and tails in ten coin tosses. Hence we are counting permutations.
So to count the favoured outcomes, which are arrangements that contain exactly $3$ heads, consider that the task is to select $3$ of $10$ 'places' to put the heads, and put tails in the remainder. That's $10$ choose $3$ ways: $^{10}\mathrm C_3$.
The number of different strings of length 5 from the alphabet $\{1,2,3,4,5\}$ is what you want to calculate. There are $5^5$ such strings, as each element of the string can be chosen in 5 different ways and you then apply the "rule of product".
However, you then mention that, "If there is a 0 in the given numbers then initial 0 is [invalid]." Now, this is not really the same question. You said you were given the alphabet $\{1,2,3,4,5\}$. If this is actually the case, you wouldn't have to worry about zeros. Then again, if you did have the alphabet $\{0,1,2,3,4,5\}$ instead, then you would have to deduct the strings of types "0xxxx","00xxx","000xx","0000x", and "00000" from $6^5$ which would be the total number of strings. I will let you figure out how many strings of type "0xxxx", etc. there are.
EDIT: The question is somewhat unclear but I notice, from the discussion in the comments above, that you are not looking for selections with repetition (?). In that case, I believe what you are looking for is a formula, also given by barak manos in the comments above, namely $\dfrac{5!}{2!2!1!}$ for "11223" for example.
To derive this, you have to realize that $5!$ is first how many ways we can permute five elements. Then, if you see the two '2':s as different, you can permute them in $2!$ ways, the '3':s in $2!$ ways, and the '1':s (trivially) in $1!$ way. The formula is then to divide the total number of permutations ($5!$ in your case) with the product of all the ways you can permute each set of elements that are "the same" (the '2':s and the '3':s in the example).
Ex. How many words can we create from "BANANA"? Answer: $\dfrac{6!}{3!2!}$ (omitting the $1!$).
Hopefully I have interpreted your question correctly, otherwise, please make it clearer.
Best Answer
Maybe, looking at an example clarifies this best :
You have $20$ objects and have to choose $5$ of them. How many possibilities are there ?
You have $20,19,18,17,16$ choices explaining how $\frac{20!}{15!}$ comes into the play.
Now each combination can appear in $5!$ possible orders which correspond to the same combination. Therefore we have to divide by $5!$ to find the number of combinations.
This can be generalized to arbitary numbers explaining how the binomial coefficient emerges.