Coin Tossing Problem: Formula for accumlating more heads than tails?

Hello

I have a 'simple' problem but having trouble working out a formula.

For example with 5 tosses & 3 Heads & 2 tails occurring, there are 5!/(3!(5-3)!) = 10 combinations.

But how many of these combinations can exist for only more or equal heads tallying to occur before the tails tally occurs.

For example

HTHTH
HTHHT
HHTTH
HHTHT
HHHTT

are all ok as the tally for H's were equal or more than the T's but

THTHH
THHTH
THHHT
TTHHH
HTTHH

is not as the tally of T's occured before H's did.

Basically if H= + 1 and T = -1 and you summed up the sequences it cannot be <0 at ANY point in the summing up.

So I need to work out the no. of outcomes from C combos of getting more heads accumulated before tails: given N no of tosses and X no of heads.

Any help in the right direction would be great.
Thanks
Max

Please read up on Bertrand's ballot theorem. You need something called reflection principle here - pretty elementary but nevertheless an ineresting application !

Brilliant thanks, thats very kind of you to respond and so quickly!