No, we can't. You can refer to the problem statement:
"A hired designer reckons that all diplomas of the same kind (for example, for winning semifinals) must be in the same row"

It's interest to solve problem if we can split. I have O(2^(3*n/2)) probably right solution using DP of bitmasks. If we have group of k diplomas where no 2 of them cannot be fully situated exactlty in one row, than we can situate these k diplomas in k or k-1 rows. For each bitmask we check, can we situate its k(mask) diplomas in k-1 rows. Than in DP for mask with k bits we check all submasks with size of <=k/2 as first such group. These get us the complicity written before.