This doesn’t give any restrictions other than both the b***s and boxes are distinguishable.
Here are solutions for certain restrictions or freedoms.
If any box(N) can contain zero to (k) b***s then the solution is
$$\displaystyle N^k \ = \ {3^6} \ = \ 729$$
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If a box MUST contain (1) but can contain up to (k-N+1) b***s then the solution is
$$\displaystyle \sum \limits_{i=0}^{\textcolor[rgb]{1,0,0}{N-1}} *(-1)^i*\binom{N}{i}* (N-i)^k \\$$
This is a scripted link to Wolfram alpha
https://www.wolframalpha.com/input/?i=sum+%28+binom%283%2Ci%29*%28-1%29^i*%283-i%29^6%29+from+i%3D0+to+2
This is the script:
sum (binom(3,i)*(-1)^i*(3-i)^6) from i=0 to 2
(Note the value of 540 is consistent with CPhill’s solutions)
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If the boxes are distinguishable and the b***s are indistinguishable then
$$\displaystyle \left( {\begin{array}{*{20}c} 56 \end{array}} \right) = \dfrac{{(N+k-1)!}}{{k!}}\; \hspace{15pt}| \hspace{15pt} \Text {N=6 \; k=3} \\$$
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When both the b***s and boxes are indistinguishable then k parts of N partitions may be needed. http://mathworld.wolfram.com/PartitionFunctionP.html This might seems easier, but this is intermediate level combinatorics. The partition function is not any more difficult to understand than combinations, but the assembly of what to add, subtract and when, increases the complexity from b***h to royal b***h.
Good thing Melody appointed me Royal SOB.
$$\tiny \textcolor[rgb]{1,,0}{\text {(Edited: Formating )}}\\$$
.
+1038 
