It might seem a bit complex to think about it initially but not really provided you imagine the ways it can happen and recognize they are all equivalent.
So this 50x happens because the chance it is found anywhere is basically the way you can have this happen out of 52 cards which is to take these 3 cards out and and then place the other 52-3=49 cards in any order you want with 1 bin separating the pieces to left and right with respect to the sequence of the 3 ABC say so that it looks like; (the separator defines where the triplet is in the sequence)

X1X2...Xk-ABC-Y1Y2..Ym

With X1,..XK|Y1...Ym now being 50 symbols (49 plus the separator) that you care for the order.

Now obviously there are 50! ways (permutations) to order these 49+1 and 49! ways to order 49 cards ignoring the bin position.

Basically there are 50! ways to order the deck with the triplet in some position intact and 49! ways to order 49 cards to begin with.

So it ought to be 50!/49! possible ways to break the sequence of 49 to left and right pieces with a separator anywhere. Or you can see is as 49! ways to orient the cards and 50 positions you can put in the ABC as triplet from left to right in a line of 49.

Those are all equivalent of course (same probability) and since the first is one of the 52C3 possible first triplets (without care for order), the chance the first triplet is the correct one is 1/52C3=1/22100~0.0045248...%

So the chance the triplet is anywhere in the deck is 50/52C3=0.22624...%


Some may simply instantly see it as 1/52C3 to grab the triplet & 50 places to put it but it needs a bit of justification properly to others.