Answer :

Make the following exchanges of pairs: H-K, H-E, H-C, H-A, I-L, I-F, I-D, K-L, G-J, J-A, F-K, L-E, D-K, E-F, E-D, E-B, B-K.
It will be found that, although the white counters can be moved to their proper places in 11 moves, if we omit all consideration of exchanges, yet the black cannot be so moved in fewer than 17 moves.
So we have to introduce waste moves with the white counters to equal the minimum required by the black.
Thus fewer than 17 moves must be impossible.
Some of the moves are, of course, interchangeable.

Math Genius