Probability of 4 heads in a row?

A coin is flipped 10 times. If it lands "heads" 4 or more times in a row, this is considered a success.
What is the probability of a success?

Actually, I already have an answer using Excel. I listed all the 1,024 ways in an Excel spreadsheet and then determined how many of those contain 4 or more heads in a row - that's 251.
So the probability of a success of 4 or more heads in a row for every 10 coin flips is 251/1,024 = 0.2451171875

Basically, I want to know the procedure for solving this type of problem (formulas - that type of thing), as opposed to working out every success out of all the possible outcomes.
(For example, if the problem had asked to find the probability of 52 or more heads in a row for every 100 flips, using Excel would be an enormous amount of work).

Thank you.

I've been searching a lot of probability websites and the best advice I've seen is: "Any probability problem never has a neat answer".

Rather than trying to solve a problem that has 1,024 outcomes, how about something smaller?

Calculating the probability of getting 3 heads (or more) in a row in five flips of a coin.
This has a total of 32 outcomes and the eight successes are:

T T H H H
T H H H T
T H H H H
H T H H H
H H H T T
H H H T H
H H H H T
H H H H H

Probability of success = 8/32 = .25

Well I think it's a good place to start (or actually to begin again).

So, from the link Denis posted I obtained these answers:
0, 0, 0, 1, 3, 8, 20, 48, 111, 251, 558, 1224, 2656, 5713
I'm guessing that the first '1' in that series is for # of successes of 4 heads in a row for 4 coin flips.
The tenth number in that series is 251 which is exactly what I calculated with an Excel spreadsheet.
Using the formula that you posted, n!/k!(n-k)!, I get an answer of 210.
So, what exactly is the formula that will generate 251?