This is one of a set of several problems in my book I am having difficulty not just solving, but also understanding the provided solutions.
The given answer is $462 - 336 + 15 = 141.$
I'll try and see where the terms of the equality above come from.
$[6] = \{1, 2, 3, 4, 5, 6\}.$
The number of all multisets of size six sourced out of $[6]$ is $\binom{6 + 6 -1}{6} = \binom{11}{6} = 462,$ so that's where the first term of $462 - 336 + 15$ come from.
All the multisets of size six where each one contains a digit that occurs at least three times can be rewritten in their type form like so: $$(3, 3), (3, 2, 1), (3, 1, 1, 1), (4, 2), (4, 1, 1), (5, 1), (6)$$
Then,
$|(3, 3)| + |(3, 2, 1)| + |(3, 1, 1, 1)| + |(4, 2)| + |(4, 1, 1)| + |(5, 1)| + |(6)| \\ = (6 \cdot 5) + (6 \cdot 5 \cdot 4) + (6 \cdot \binom 53) + (6 \cdot 5) + (6 \cdot \binom 52) + (6 \cdot 5) + 6 \\ = 30 + 120 + 60 + 30 + 60 + 30 + 6 \\ = 336.$
So that's where the second term of $462 - 336 + 15$ must come from.
I am not clear on where $15$ in $462 - 336 + 15$ come from. So, that's what I'd like to know. Thanks.