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Give combinatorial a proof that $$\binom{n}{2}^2=\binom{n}{2}+6\binom{n}{3}+6\binom{n}{4}$$

I firstly thought it like classical commission questions consisting of $n$ men and $n$ women etc. However , i did not work. After that , i thought about manipulating the question such that $$\binom{n}{2}^2=\binom{n}{2}+6\binom{n}{3}+6\binom{n}{4}$$ $$\binom{n}{2}\binom{n}{n-2}=\binom{n}{2}+6\binom{n+1}{4}$$ However , i cannot still find anything. Can you help me ?

NOTE: Although @Phicar's answer get credit from you , i could not comprehend it :( So , i am open to other proofs.

NOTE 2: Do not write "HINTS" , if you want to write somethings , please be clear and elaborately

3 Answers3

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Hint: Try the most natural thing: You are choosing ordered pairs of groups of 2 people out of $n$ people, this is the left hand side. You can either have the same group, that is, the pair share both people (in how many ways?) or you choosed two groups that share exactly one person (so you are choosing $3$ people, actually) or your pair has no one in common (so you are choosing 4 people).

Edit: You are asking for not a hint..well consider $\binom{[n]}{k}=\{A\subseteq [n]=\{1,2,\cdots ,n\}:|A|=k\},$ what you want is to find a bijection in between $\binom{[n]}{2}\times \binom{[n]}{2}$ and $\binom{[n]}{2}\bigcup \binom{[n]}{3}\times \binom{[3]}{1}\times\binom{[2]}{1} \bigcup \binom{[n]}{4}\times\binom{[4]}{2}$, where those are disjoint unions. For that, consider the following function $$\varphi (A,B)=\begin{cases}A & \text{if }A=B\\ (A\cup B,A\cap B,A\setminus B) & \text{ if }|A\cap B|=1\\ (A\cup B,A) & \text{Otherwise, so}|A\cap B|=0\end{cases},$$ Just show that this is actually a bijection.

Phicar
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  • your answer did not help me , i cannot see it. You said that "$2$ people out of $n$ people" , but L.H.S is $C(n,2)^2$. –  Jul 29 '22 at 03:12
  • I did not understand the "ordered pairs" example –  Jul 29 '22 at 03:16
  • @HakunaMatata I said ordered pairs of 2 people out of $n$ people, meaning by product rule $C(n,2)^2$. – Phicar Jul 29 '22 at 03:18
  • @HakunaMatata Ordered pair meaning $(a,b)$ where $a,b$ are sets of size $2$. – Phicar Jul 29 '22 at 03:20
  • if we pick two pairs from a single group of $n$ people , then it must be $C(n,2)\times C(n-2,2)$ , not $C(n,2)\times C(n,2)$. I really cannot understand the left handside –  Jul 29 '22 at 09:40
  • @HakunaMatata you could pick the same group or groups that shared a person, so as to make the equation – MafPrivate Jul 29 '22 at 10:20
  • @HakunaMatata I have edited with the more details of the approach, more like what a formal proof of what I mean is. – Phicar Jul 29 '22 at 16:03
  • Hopefully clear but longer rewording of Phicar's answer: Consider a two round lucky draw contest which selects two people out of $n$ for both the rounds so that a total of $4$ people are selected. Directly this can be done by selecting 2 people for the first round which can be done in $C(n,2)$ ways AND then again selecting 2 people for the 2nd round which can be done in another $C(n,2)$ ways. We need to multiply these two because of the logical "AND" (AND=multiply, OR=add). So the LHS is $C(n,2)^2$ ways. – Sanjana Dec 08 '23 at 16:37
  • [Continued...] For the RHS we select again 4 people but in a different way...We select 4 people for both the rounds together and redistribute them. This can be done only in the following three possible ways---
    1. The two people selected for first round are reselected for the 2nd i.e. instead of 4, only 2 are selected twice. This can happen in $C(n,2)$ ways because the selection in first

    round determines that in 2nd.

     – Sanjana Dec 08 '23 at 16:39
  • [Continued...] 2. One out of the two people selected for 1st round is reselected. So overall 3 people are selected out of $n$. In how many ways can this happen? First select the 3 people from $n$, which can be done in $C(n,3)$ ways AND then out of the chosen 3 select that person who is to be reselected which can be done in $C(3,1)$ ways AND then you place the 2 people in the remaining 2 places(2 because there were total 4 places and 2 are occupied by the guy who was chosen in both rounds)---this can be done in 2! ways. – Sanjana Dec 08 '23 at 16:40
  • [Continued...] 3. Finally comes the case where all 4 people are different i.e. none are reselected. So you actually choose 4 people out of $n$ but then you can choose 2 out of them in the 1st round in $C(4,2)$ ways and then you can put the rest of them together in the 2nd round in only one way (ordering doesn't matter!)

    Add the results up: that's the RHS.

     – Sanjana Dec 08 '23 at 16:41
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The other answer by Phicar is splendid, but I thought I should relate it more concretely to your attempt, picking two (unordered, but distinguishable) pairs from a group

consisting of $n$ men and $n$ women

Instead, try to pick two (not necessarily disjoint) pairs from a single group of $n$ people. The left-hand side is the direct approach, while the right-hand side splits into cases based on how many people were actually chosen.

Arthur
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  • if we pick two pairs from a single group of $n$ people , then it must be $C(n,2)\times C(n-2,2)$ , not $C(n,2)\times C(n,2)$. I really cannot understand the left handside –  Jul 29 '22 at 09:40
  • @HakunaMatata The two pairs we pick do not have to be disjoint. Maybe the first pair has Alice and Bob, and the second pair has Alice and Charlie. That's entirely allowed. As for the right-hand side, well, in my example here I happened to pick three people: Alice, Bob, and Charlie. Picking 3 people can be done in $\binom n3$ ways. Once we've picked three people, there are six ways to make up the two pairs (remember, $\binom n2^2$ says it matters which pair is first, but not which person is first in a given pair):$$AB, AC\AC, AB\AB, BC\BC, AB\AC, BC\BC, AC$$Do the same for picking 2 or 4. – Arthur Jul 29 '22 at 09:44
  • but the selection order does not matter , right ? For example if alice and bob are in the same pair , it does not matter whether Alice or bob selected first. The only matter is the order of pairs .For example ,selection order of alice and bob and alice and charlie matter. Am i right ? –  Jul 29 '22 at 09:54
  • As long as we know Alice and Bob are in the first pair, it does not matter which order they are picked for that pair, no. If you want a real-world example, consider a meeting with $n$ people where you choose two people to bring (identical) cakes, and during the meeting you have a lottery where two people win a cake. Where those who brought cake can win it back if they're lucky. The right-hand side is then splitting into cases depending on how many bakers got to bring their cake back home (but deciding who bakes and who wins after choosing which two-to-four people get picked at all). – Arthur Jul 29 '22 at 11:57
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It might help to explain the origin of the two sixes since they come from different origins.

The green part below consists of choosing one of three to be doubled, multiplied by the ways of choosing which of the other two goes where. The blue part is the number of ways we can split four into two pairs. $$\binom{n}{2}^2=\color{purple}{\binom{n}{2}}+\color{green}{\binom{3}{1}\binom{2}{1}\binom{n}{3}}+\color{blue}{\binom{4}{2}\binom{n}{4}}$$ This is not a combinatoric proof, but just to confirm the answer: $$ \frac12n(n-1)+6\frac16n(n-1)(n-2)+6\frac1{24}n(n-1)(n-2)(n-3)\\ =\frac14n(n-1)\left\{2+4(n-2)+(n-2)(n-3)\right\}\\ =\frac14n(n-1)\left\{n(n-1)\right\}\\ =\binom{n}{2}^2 $$

Suzu Hirose
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