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Mar 6 at 18:01 comment added MikeTeX Very nice and instructive answer.
Mar 6 at 17:59 comment added MikeTeX N(j) = $\sum_{n\geq j} a_n {n \choose j}$, where $a_n$ is the number of elements that are exactly in $n$ of the $S(i)$. Then the cancelation lemma implies that $\sum_{j \geq k} (-1)^{j-k} {j \choose k} N(j) = a_k$.
Mar 6 at 17:59 comment added MikeTeX OK. Note: I would write your proof of (6) as follows:
Mar 6 at 17:52 comment added robjohn For $0\le j\lt k$, $\binom{j}{k}=0$, so starting at $0$ or $k$ gives the same result. My personal preference is to start at $0$ as that often proves easier in cases where the order of summation needs to be changed.
Mar 6 at 17:37 comment added MikeTeX Typo: I think in (6), the index j should start from k.
Aug 30, 2021 at 3:10 history bounty ended linear_combinatori_probabi
Sep 3, 2020 at 3:51 history bounty ended linear_combinatori_probabi
Oct 29, 2019 at 7:56 history edited robjohn CC BY-SA 4.0
improve the explanation of $\binom{-1}{n}$
Oct 28, 2019 at 22:42 history edited robjohn CC BY-SA 4.0
add Corollary 2
Oct 16, 2019 at 2:34 history edited robjohn CC BY-SA 4.0
color the theorem and corollary the same as the lemma
Oct 15, 2019 at 23:00 history edited robjohn CC BY-SA 4.0
fix a minor point
Oct 15, 2019 at 19:43 history edited darij grinberg CC BY-SA 4.0
added 2 characters in body
Oct 15, 2019 at 19:04 history edited darij grinberg CC BY-SA 4.0
the factorial formula requires n >= k
Oct 15, 2019 at 18:59 history edited darij grinberg CC BY-SA 4.0
the factorial formula requires n >= k
Oct 5, 2019 at 13:24 history edited robjohn CC BY-SA 4.0
improve the subscripts for the underbraces
Jul 12, 2017 at 15:21 history edited robjohn CC BY-SA 3.0
Add $(5)$
Sep 28, 2016 at 13:28 history edited robjohn CC BY-SA 3.0
improve underbrace
Sep 11, 2013 at 23:08 history edited robjohn CC BY-SA 3.0
fix typo
Jul 6, 2013 at 14:43 history edited robjohn CC BY-SA 3.0
add a corollary
Jun 11, 2013 at 0:09 history edited robjohn CC BY-SA 3.0
renumber
Jun 11, 2013 at 0:03 history edited robjohn CC BY-SA 3.0
add the simplest proof of the cancellation lemma
Apr 15, 2013 at 21:48 history edited robjohn CC BY-SA 3.0
add a second proof
Apr 15, 2013 at 16:27 history answered robjohn CC BY-SA 3.0