CareerCup

A classic question is: Divide into two sets, such that each contains the same number of heads (note: not necessarily 10 heads each).

Perhaps you have your own interpretation in mind and didn't really understand the actual question?

//can not be done exactly. gut here is the try ->
flip 10 of them ,, not we have majority of heads or tails except we got all heads or all tails . move 10 of them . flip the other group of coins.

flip 10 of them ,, now we have majority of heads or tails ,except we gets all heads or all tails . move 10 of them to a group . flip the other group of coins.

For any given number, divide them into two group of equal number.
Say we have 20 Coins, and we have to divide them into 10-10's Group,

First 10 would contain say x Heads, and ( 10 - x ) tails.
while other group would contain ( 10 - x ) heads and x Tails.

Just flip one Group totally, and it would be same as the other Group. in other words,
x heads and ( 10 - x ) tails would convert into ( 10 - x ) heads and x tails. which is similar to other group.

I tried solving the question as you phrased it but no luck, as far as I can tell it looks impossible.
does anyone think there is a way to actually separate the coins into two setts of either all head or all tail?