Same Heads

My friend sent me this:
"There are 42 coins on a table. Of those, 10 are heads-up and the other 32 are tails-up.

How can you separate the coins into two groups so that each contains the same number of coins that are heads-up?

Note that you cannot see nor feel the coins, though you can manipulate them. (you are blindfolded and wearing gloves)."

I solved it and proved that my solution is correct. I'm interested in seeing if other solutions exist.

@damionhonegan said
My friend sent me this:
"There are 42 coins on a table. Of those, 10 are heads-up and the other 32 are tails-up.

How can you separate the coins into two groups so that each contains the same number of coins that are heads-up?

Note that you cannot see nor feel the coins, though you can manipulate them. (you are blindfolded and wearing gloves)."

I solved it and proved that my solution is correct. I'm interested in seeing if other solutions exist.

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@forkedknight

This is my solution. Are there any other?

@damionhonegan said
This is my solution. Are there any other?

Put 32 coins into a new pile and flip the ten coins left in the other pile.

Put all the coins on the edge.

More seriously
If there are H heads
Two piles of coins
And all coins in one pile, size P, with h heads in it, are flipped
(You have to flip all coins in a pile. If you don’t, an unflipped coin could be heads or tails, which changes the count in that pile independently of the other pile.)

Before flip there were h and H-h heads in each pile
After the flip there are P-h and H-h heads in each pile

For the piles to have equal numbers of heads
P-h = H-h
P = H

So the size of the pile flipped must be equal to the total number of heads