Use flavour multiplicities rather than physical bar identities

For n different questions, there are $n!$ orders. Since $4!=24<25$ and $5!=120\ge25$, the least n is 5.

The source table is white:4, milk:3, dark:4, hazelnut:1. Friends are distinct recipients; same-flavour bars are treated as indistinguishable. Any non-hazelnut flavour can supply all three friends, but hazelnut can go to at most one.

All distinct flavours: ${}_4P_3=24$. All the same: $\binom31=3$, since only the three non-hazelnut flavours have enough bars. For exactly one repeated flavour, choose the repeated non-hazelnut flavour in 3 ways, a different flavour in 3 ways, then its position pattern in $3!/2!$ ways:27. The source writes this equal count as $\binom31\binom32\,3!/2!=27$; the second factor is numerically 3, the available different-flavour choices. Total $24+3+27=54$.

If no friend gets hazelnut, each gets one of 3 flavours: $3^3=27$. If one gets hazelnut, choose that recipient in $\binom31$ ways and the other two flavours independently in 3 ways each: $\binom31\times3\times3=27$. Total 54.

If the particular friend receives hazelnut, the other two receive different non-hazelnut flavours in $\binom32\,2!$ ways or the same flavour in $\binom31$ ways. This gives $6+3=9$ forbidden assignments. Hence $54-9=45$.

No hazelnut gives 27. Hazelnut given to one of the other two friends gives $\binom21\times3\times3=18$. Thus $27+18=45$.