These were simple induction proofs, so I decided to try and prove them combinatorially. I think I nailed the first one, not so sure about the second one.


Have $n+1$ integers want all orderings except for strictly ascending. Can pick all permutations and subtract one(right hand side) or remove each # $\ge$ minimum # one at a time and place to the left of the ordering, gauranteeing they will never be in strictly ascending order(left hand side). There would be $n$ of these #'s times $n!$ for the ordering of the remaining #'s. $(i)(i!)$ 'occurs' $n$ times making it $n(n!)$.


Have 2 groups of $n$ bits each. want # combinations of which bit(they are numbered/specified) selected from first group with all configurations of bits in $2nd$ group(they are not numbered/specified). LHS $i(2^i)$ 'occurs' $n$ times making it $n(2^n)$. This is how the bits are first placed. For RHS can remove one bit from first group and place it in the $2nd$ so that new total is the configuration$(1/0)$ of bit selected$(2)$ plus all combos of remaining bits in first group$(n-1)$ with all configs of bits in second group(now $2^{n+1}$).

Please verify proofs, especially second one. Thanks.

  • 2
    $\begingroup$ Are you asking for us to check your proofs? Your question isn't entirely clear. If that is what you want, then I suggest the proof-verification tag. $\endgroup$ Oct 25 '14 at 18:28
  • $\begingroup$ As was obvious to begin with and has now been made explicit, the OP is asking for verification of the proofs. $\endgroup$ Oct 26 '14 at 2:56
  • $\begingroup$ Update: Thanks for the replies. Awesome. My attempts make me feel like an @ss, though. lol. $\endgroup$
    – miniparser
    Mar 14 '15 at 21:48

In the first one the lefthand side isn’t $$\underbrace{n!+n!+\ldots+n!}_{n\text{ times}}\;,$$ which is what you’ve described, but rather $$1\cdot1!+2\cdot2!+\ldots+n\cdot n!\;.$$ I can’t altogether follow your argument for the second one, but I think that you’ve made a similar mistake in interpreting the summation.

I’ve not yet come up with a combinatorial argument for the first identity, and I don’t really expect to, but I have one for the second. You have $n+1$ slips of paper, numbered $0$ through $n$, lined up on the table, white on one side and red on the other. All of them are white side up. Now choose a number $k\in\{1,\ldots,n\}$, and turn over slip $k$. Then turn over any subset of the $k$ slips to the left of slip $k$ (i.e., though with numbers less than $k$); you may choose to turn over none at all. Finally, choose any one of the $k$ slips to the left of slip $k$ and put a coin on it; it doesn’t matter whether the slip is white side up or red side up. For a fixed choice of $k$ there are $k2^k$ ways to do this, so there are $\sum_{k=1}^nk2^k$ ways altogether, and they all produce different outcomes.

Now imagine that instead you turn over any subset of the $n+1$ slips and put a coin on any slip except slip $n$. If there is no red slip to the right of the slip with the coin, turn over slip $n+1$. There are altogether $n2^{n+1}$ ways to do this, and the possible outcomes are exactly the same as for the first procedure. However, the outcomes in which slip $n+1$ is the only red slip to the right of the coin is produced in two ways, one in which it was part of the set of slips originally chosen to be turned over, and one in which it was not. I claim that there are $2(2^n-1)$ such outcomes. If this is correct, the number of distinct outcomes is $n2^{n+1}-2(2^n-1)=(n-1)2^{n+1}+2$, and the identity is established.

Consider an outcome in which slip $n+1$ is the only red slip to the right of the coin. Let $S$ be the set of slips that either are red or have the coin; $S$ can be any non-empty subset of the first $n$ slips, so there are $2^n-1$ possibilities for $S$. Let $k$ be the number of the slip with the coin. Then $k=\max S$, and there are two possibilities: either slip $k$ is red, or it isn’t. Thus, there are indeed $2(2^n-1)$ outcomes of this type, and we’re done.

  • $\begingroup$ If that is the case for the first one I have trouble reconciling the answer given for this one: math.stackexchange.com/questions/950923/… as it claims there are n light switches not $1+2+3+...+n$. $\endgroup$
    – miniparser
    Oct 26 '14 at 20:25
  • $\begingroup$ @user2008730: I don’t understand what problem you’re having, I’m afraid. Can you be a bit more explicit? $\endgroup$ Oct 26 '14 at 20:28
  • $\begingroup$ In light of what's been said in response to this posting, I don't think the answer to the linked problem is correct. $\endgroup$
    – miniparser
    Oct 26 '14 at 23:16
  • $\begingroup$ @user2008730: It is correct, and nothing that I’ve said here has any real bearing on that problem. Will Orrick’s combinatorial argument for the first identity, on the other hand, uses essentially the same idea as the solution to the linked problem. $\endgroup$ Oct 26 '14 at 23:18
  • $\begingroup$ I know how to interpret a summation, that's pretty elementary. It just seems that in combinatorial proofs it is not always interpreted literally, as in the linked problem. $\endgroup$
    – miniparser
    Oct 27 '14 at 1:11

For the first identity, you seem to be on the right track, but some things in your description are unclear to me and the sentence “$(i)(i!)$ 'occurs' $n$ times making it $n(n!)$” isn't correct.

As you suggest, the righthand side counts non-identity permutations of $(1,2,3,\ldots,n+1).$ So does the lefthand side: each non-identity permutation has at least one element that is not in its original position. Let $S_i$ be the set of permutations in which elements $1,2,\ldots,n-i$ are in their original positions, but element $n+1-i$ is not. There are $i$ elements that can be placed in position $n+1-i$ (any of elements $n+2-i$ through $n+1$). Then there remain $i$ elements that can be placed in any order in positions $n+2-i$ through $n+1.$ Now sum the sizes of $S_i$ for $i$ from $0$ to $n.$ Note that $S_0$ is empty and can therefore be omitted from the sum.

Added: For the second identity, I provide the following alternative to Brian Scott's proof. This proof avoids the double counting of certain outcomes and consequent need for subtraction of Brian Scott's proof, but is otherwise similar.

Say we have $n+1$ people ranked according to seniority from $0$ to $n,$ with $n$ being most senior. We are to form a committee consisting of a chair and zero or more additional members. The chair must be the most senior member of the committee. In addition, we are to choose a spokesperson who may or may not also serve on the committee, and who must be less senior than the chair. This interpretation is completely isomorphic to Brian Scott's interpretation involving colored slips of paper and a coin; I offer it only for variety and because its the one I came up with when I started thinking about the problem.

The number of ways of selecting chair, remaining committee members, and spokesperson is $\sum_{i=1}^n\,i\,2^i$ by the argument in Brian Scott's proof. We represent such a selection by a triple $(c,M,s),$ where $c$ is the chair, $M$ is the set of other committee members, and $s$ is the spokesperson. Observe that there are two selections in which the chair has rank $1$, namely $(1,\{\},0)$ and $(1,\{0\},0).$

To establish the identity, we exhibit an alternate procedure for making the selection in the case that the chair has rank $2$ or higher. The procedure is:

  1. choose a subset $S$ of $\{0,1,\ldots,n\};$
  2. choose a rank $p$ between $1$ and $n-1;$
  3. if $S$ contains a person of rank greater than $p,$ let $c$ be the highest ranked element of $S,$ let $M=S\setminus\{c\},$ and let $s=p;$
  4. if $S$ contains no person of rank greater than $p,$ let $c=p+1,$ let $M=S,$ and let $s=0.$

It is clear that this procedure always produces a valid selection, $(c,M,s)$ with $c\ge2.$ Furthermore, any valid selection, $(c,M,s),$ with $c\ge2$ is produced in a unique way by this procedure: if $s>0,$ then $(c,M,s)$ arises when $S=\{c\}\cup M$ and $p=s;$ if $s=0,$ then $(c,M,s)$ arises when $S=M$ and $p=c-1.$ It is clear that the selection can be accomplished in $(n-1)2^{n+1}$ ways.

Visual proof: I would suggest modifying $$ \sum_{i=1}^n i\,2^i=2+(n-1)2^{n+1} $$ by dropping the first term in the sum on the left and the term $2$ on the right. Then replace $n$ with $n+1$ to get $$ \sum_{i=2}^{n+1} i\,2^i=n\,2^{n+2}. $$ Divide both sides by $2^2$ and reindex the sum to obtain the equivalent identity, $$ \sum_{i=1}^n (i+1)\,2^{i-1}=n\,2^n, $$ which has the visual proof shown below.

enter image description here

Relationship between the visual proof and the preceding bijective proof: We can see the diagram as a proof that $$ \sum_{i=2}^5i\,2^i=4\cdot2^6. $$ Imagine the rectangle divided into $2^6=64$ columns. So the light gray rectangle in the upper left would get divided into four columns, as would the adjacent light gray rectangle; the next, slightly darker gray rectangle would get divided into eight columns, and so on. Label columns from left to right by subsets $S\subseteq\{0,1,2,\ldots,5\}$ using the natural ordering $\{\},$ $\{0\},$ $\{1\},$ $\{0,1\},$ $\{2\},$ $\{0,2\},$ $\{1,2\},$ $\{0,1,2\},$ $\{3\},\ \ldots.$ Label rows from top to bottom by integers $p=1,2,3,4.$

So the column labels $\{\},$ $\{0\},$ $\{1\},$ $\{0,1\}$ would appear above the first light gray block, the labels $\{2\},$ $\{0,2\},$ $\{1,2\}$ above the second light gray block, the labels $\{3\},\ \ldots$, $\{0,1,2,3\}$ above the next, slightly darker gray block, and so on.

The pairs $(S,p)$ to which rule $4$ in the bijection above applies fall within the first block of each row of the diagram; the pairs to which rule $3$ applies all fall within the later blocks of each row.

Historical note: There is a visual proof by Oresme (see Victor J. Katz's A History of Mathematics: An Introduction, Third Edition, page 358) that $$ 1\cdot\frac{1}{2}+2\cdot\frac{1}{4}+3\cdot\frac{1}{8}+4\cdot\frac{1}{16}+\ldots=2, $$ which seems similar in some respects to the visual proof above.

Some thoughts: The notion of combinatorial proof can be rather slippery. I generally interpret it to mean bijective proof, which is proof in which the two sides of an identity are interpreted as counting two sets of objects, and then those sets of objects are shown to be equinumerous by exhibiting a bijection between them, thereby establishing the identity. A special case is double-counting proof, in which a single set of objects is counted in two different ways. This is bijective proof in which the bijection is the identity.

The step in which the each side of the identity is interpreted as counting something is key. The interpretation should be natural, making use of straightforward counting principles.

I consider the proof above involving the committee to be combinatorial: the left side counts a specified set of triples $(c,M,s)$ in a straightforward way, while the right side counts a specified set of ordered pairs $(S,p)$ in an even more straightforward way. Finally, there is a bijection between the triples and the pairs.

The visual proof is also combinatorial in this sense—in some ways, it is even better. It is a double-counting proof in which the rectangle area (which consists of unit squares) is computed in two different ways—on the left by adding L-shaped regions, on the right by the trivial area formula. In another sense, however, the visual proof is not combinatorial. The exponential function $2^n$ is not given combinatorial meaning (say as the size of the power set). Only an arithmetical property is used, namely that $2^n=2^{n-1}+2^{n-1}.$

  • $\begingroup$ In the post you link to, consider $n=3.$ Then $\sum_{i=1}^n2^{i-1}=2^0+2^1+2^2=1+2+4=7,$ while $2^n-1=8-1=7.$ The patterns with at least one light on are $100,010,110,001,101,011,111.$ Notice that the first one has the first light on and all subsequent lights off. The next two have the second light on and the subsequent light off. The last four have the third light on. (There are no subsequent lights in this case, but it is vacuously true that all lights to the right of the third light are off.) This provides a combinatorial interpretation of the lefthand side. $\endgroup$ Oct 27 '14 at 3:29

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