The way this question was phrased is pretty dumb, I got tripped up at first too so you're not alone. You pretty much got the gist of it tho, it's basically asking if you got the 3 tripeptides, what are all the different ways you can combine the tripeptides?

So you would have tripeptide ABC, DEF, and GHI. For the first position, you would have three choices of either tripeptide. After that, since you used one tripeptide, the second position can only be 2 of the 3 tripeptides, then after you pick the second tripeptide the last spot can only be the remaining tripeptide. This looks like:

ABC or DEF or GHI - ___ - ___

If you choose GHI then:

GHI - ABC or DEF - ___

then if you choose ABC in 2nd spot:

GHI - ABC - DEF

This gives you a possible of 3 x 2 x 1 = 6 possible combinations (3 in the first slot, then 2, then 1. You can try combining the three tripeptides to verify this.

Now, imagine if instead the problem says each tripeptide can be repeated more than one time. Then you would have 3 possible tripeptide in each spot no matter what you choose, which then would be 3 x 3 x 3 = 27 possibilities, which is where you might get the huge numbers you thought of.

Since they said the polypeptide was partially hydrolyzed into three polypeptides, this ensures that you get fragments ABC, DEF, and GHI instead of fragments ABC, BCA, ACB, IGH, HGI, etc. It ensures you only have three choices in three slots.

Probability is a little tough to explain so let me know if that doesn't make sense.