We don't understand what you mean by "different marks".
If you meant three cards like these, different ranks and different suits:
J♥, 3♦♦, 8♣,
then there are 13C3 ways to choose the 3 ranks, and for each of those ways,
there are 4 suits to give the lowest-ranked card, and for each of those ways,
there are 3 suits remaining to give the card with the middle-rank, and
2 suits remaining to give to the highest-ranked card.
The numerator of the probability would be (13C3)(4)(3)(2) = 286*24 = 6864.
The denominator would be the number of ways to select any 3 cards, or 52C3,
or 22100.
That would be a probability of 
which reduces to 
,
not 
.
So we must be misinterpreting what you mean by "different marks".
Sorry.
Edwin
