The number of bags of k N-bit numbers is 2^N + k - 1 choose k: https://en.wikipedia.org/wiki/Multiset#Counting_multisets https://en.wikipedia.org/wiki/Multiset

The number of bags of k N-bit numbers is 2^N + k - 1 choose k: https://en.wikipedia.org/wiki/Multiset#Counting_multisets https://en.wikipedia.org/wiki/Multiset#Counting_multisets The number of bits required to uniquely represent such a bag is thus ⌈log2((2^N+k-1) choose k)⌉. For (N, k) we have: (6, 6) -> ⌈log2(69 choose 6)⌉ = ⌈log2(119877472)⌉ = 27 bits (6, 7) -> ⌈log2(70 choose 7)⌉ = ⌈log2(1198774720)⌉ = 31 bits So you can store seven 6-bit integers without order in one int32. I don't know if it'll be slow, though. It should be possible to infer an encoding algorithm from the proof of the counting formula but it might be slow.