<p>The formula you wrote comes from the mathematical concept of <strong>combination.</strong></p>
<p>In mathematics, a <em>combination</em> is a selection of items from a set that has distinct members, such that the <strong>order of selection does not matter</strong>: you take a certain number of objects from a set, without caring of the order of the selection.</p>
<p>The <strong>number of combinations of <span class="math-container">$n$</span> things taken <span class="math-container">$k$</span> at a time,</strong> without repetition (each element is taken only once), can be indicated as <span class="math-container">$C^n_k$</span>, and it is equal to the so-called <em>binomial coefficient</em>, indicated as <span class="math-container">$\binom{n}{k}$</span>:</p>
<p><span class="math-container">$$C^n_k=\binom{n}{k}\equiv \frac {n!}{k!(n-k)!}$$</span></p>
<p>In our case, of an <strong>exchange economy with <span class="math-container">$n$</span> goods,</strong> the number of markets (the exchange spots of two goods) is given by the combinations of <span class="math-container">$n$</span> objects taken two at a time, that is:</p>
<p><span class="math-container">$\;$</span></p>
<p><span class="math-container">$$C^n_2= \binom{n}{2}=\frac {n!}{2!(n-2)!}= \frac {n(n-1)(n-2)(n-3)... 2\cdot 1}{2(n-2)(n-3)... 2 \cdot 1}=\frac {n(n-1)}{2}\qquad (1)$$</span>
<span class="math-container">$\;$</span></p>
<p>If you consider relative prices regardless of the order of the goods, that is, given two goods <span class="math-container">$A$</span> and <span class="math-container">$B$</span>, you consider <span class="math-container">$\frac{p_A}{p_B}$</span> and <span class="math-container">$\frac{p_B}{p_A}$</span> to be the same price, the number of relative prices is also given by the formula <span class="math-container">$(1)$</span> (<span class="math-container">$p_A$</span> and <span class="math-container">$p_B$</span> are the nominal prices of <span class="math-container">$A$</span> and <span class="math-container">$B$</span>).</p>
<p><span class="math-container">$\;$</span></p>
<p>You can see:</p>
<p><a href="https://en.wikipedia.org/wiki/Combination" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Combination</a></p>
<p><a href="https://en.wikipedia.org/wiki/Binomial_coefficient" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Binomial_coefficient</a></p>

CC BY-SA 4.0
2023-07-04T12:39:24.917Z
Thanks a lot. Very clear explenation!
BlankerHans
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2023-07-04T12:46:09.680Z
You are welcome!
BakerStreet
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2023-07-04T13:13:13.920Z
I do have one more question: If we consider an exchange economy with $n$ goods and $one$ universal used medium of exchange in which every other good is priced can we formulate it also with the binomial coefficient (what would $k$ be in this case)? Intuitively I get that it is $n-1$ because every other good is priced in that specific good beside the good it self.
BlankerHans
CC BY-SA 4.0
2023-07-04T13:30:56.520Z
I agree with you, if you are saying that the only prices you consider are $\frac{p_i}{p_U}$, $i=1,...,n-1$ (the goods except the universal medium),   where $p_U$ is the price (arbitrary) of the 'universal medium of exchange'.
BakerStreet
CC BY-SA 4.0
2023-07-05T15:56:44.927Z
Now I am interested in a way to proof that $(n(n-1))/2 > n-1$ holds true. I would like to make the case that an economy with one universal medium of exchange is more efficient then a pure barter economy.
BlankerHans
CC BY-SA 4.0
2023-07-05T16:09:31.500Z
This true for $n>2$. 
Indeed, for $n\neq 1$ we have, (dividing both sides by $n-1$),  $\frac {n(n-1)}{2}>n-1  \iff n/2>1 \iff n>2$
BakerStreet
CC BY-SA 4.0
2023-07-05T19:27:06.237Z
Oh now its obvious for me. Thank you very much! I think this is all. Have a nice day :)
BlankerHans
CC BY-SA 4.0
2023-07-05T20:02:13.770Z
You are welcome:)
BakerStreet
CC BY-SA 4.0
2023-07-04T12:23:33.877Z
2023-07-27T00:23:58.857Z
2023-07-27T00:23:58.857Z
BakerStreet
<p>So recently I am reading a lot of monetary literature and this comes up when they talk about the unit of account function of money:</p>
<p><span class="math-container">$$(n / 2)(n-1)$$</span></p>
<p>Can we prove this or where does it come from and what is the intuition here?</p>

CC BY-SA 4.0
2023-07-04T01:11:27.163Z
What are you reading exactly? please clarify this question
EconJohn
CC BY-SA 4.0
2023-07-04T11:48:56.150Z
Bofinger, P., 2001. Monetary policy: goals, institutions, strategies, and instruments. and Issing, O., 2011. Einführung in die Geldtheorie. for example mention this formula.
BlankerHans
<money><price>
The number of relative prices without money
BlankerHans
# The number of relative prices without money

(Asked by: username_0 on 2023-07-03)

So recently I am reading a lot of monetary literature and this comes up when they talk about the unit of account function of money:

$$(n / 2)(n-1)$$

Can we prove this or where does it come from and what is the intuition here?

**Tags:** money, price

> **Comments:**
> **username_1**: What are you reading exactly? please clarify this question  
> **username_0**: Bofinger, P., 2001. Monetary policy: goals, institutions, strategies, and instruments. and Issing, O., 2011. Einführung in die Geldtheorie. for example mention this formula.  

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## 1 answers

### Answer by username_2 on 2023-07-04. Score: 4 | Accepted answer

The formula you wrote comes from the mathematical concept of **combination.**

In mathematics, a *combination* is a selection of items from a set that has distinct members, such that the **order of selection does not matter**: you take a certain number of objects from a set, without caring of the order of the selection.

The **number of combinations of $n$ things taken $k$ at a time,** without repetition (each element is taken only once), can be indicated as $C^n\_k$, and it is equal to the so-called *binomial coefficient*, indicated as $\binom{n}{k}$:

$$C^n\_k=\binom{n}{k}\equiv \frac {n!}{k!(n-k)!}$$

In our case, of an **exchange economy with $n$ goods,** the number of markets (the exchange spots of two goods) is given by the combinations of $n$ objects taken two at a time, that is:

$\;$

$$C^n\_2= \binom{n}{2}=\frac {n!}{2!(n-2)!}= \frac {n(n-1)(n-2)(n-3)... 2\cdot 1}{2(n-2)(n-3)... 2 \cdot 1}=\frac {n(n-1)}{2}\qquad (1)$$
$\;$

If you consider relative prices regardless of the order of the goods, that is, given two goods $A$ and $B$, you consider $\frac{p\_A}{p\_B}$ and $\frac{p\_B}{p\_A}$ to be the same price, the number of relative prices is also given by the formula $(1)$ ($p\_A$ and $p\_B$ are the nominal prices of $A$ and $B$).

$\;$

You can see:

<https://en.wikipedia.org/wiki/Combination>

<https://en.wikipedia.org/wiki/Binomial_coefficient>

> **Comments:**
> **username_0**: Thanks a lot. Very clear explenation!  
> **username_2**: You are welcome!  
> **username_0**: I do have one more question: If we consider an exchange economy with $n$ goods and $one$ universal used medium of exchange in which every other good is priced can we formulate it also with the binomial coefficient (what would $k$ be in this case)? Intuitively I get that it is $n-1$ because every other good is priced in that specific good beside the good it self.  
> **username_2**: I agree with you, if you are saying that the only prices you consider are $\frac{p_i}{p_U}$, $i=1,...,n-1$ (the goods except the universal medium),   where $p_U$ is the price (arbitrary) of the 'universal medium of exchange'.  
> **username_0**: Now I am interested in a way to proof that $(n(n-1))/2 > n-1$ holds true. I would like to make the case that an economy with one universal medium of exchange is more efficient then a pure barter economy.  
> **username_2**: This true for $n>2$.  Indeed, for $n\neq 1$ we have, (dividing both sides by $n-1$),  $\frac {n(n-1)}{2}>n-1  \iff n/2>1 \iff n>2$  
> **username_0**: Oh now its obvious for me. Thank you very much! I think this is all. Have a nice day :)  
> **username_2**: You are welcome:)  

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