ch1.2.11-transpositions-and-binomial-proof

Ch 1.2.11 — Of the Transposition of the Letters

Summary: Proves the binomial coefficient formula by counting letter permutations; introduces factorials and the multinomial coefficient, and extends the result to powers of trinomials.

Sources: chapter-1.2.11

Last updated: 2026-04-29

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Coefficients as Permutation Counts (article 352)

Euler explains why the binomial coefficients have the values they do. When $(a+b{)}^n$ is expanded by multiplying $n$ factors of $(a+b)$, each term $a^{n-k}{b}^k$ arises from choosing $b$ from exactly $k$ of the $n$ factors. The term therefore appears once for every distinct ordering of the letters $\underset{n-k}{\underbrace{a\cdots a}}\underset{k}{\underbrace{b\cdots b}}$.

Examples from the 4th power:

• $a^{3}b=aaab$: 4 orderings ($aaab,aaba,abaa,baaa$) → coefficient 4.
• $a^2{b}^2=aabb$: 6 orderings ($aabb,abba,baba,abab,bbaa,baab$) → coefficient 6. (source: chapter-1.2.11)

Factorials: Permutations of $n$ Distinct Letters (articles 355–356)

For $n$ mutually distinct letters, the number of orderings (permutations) grows as the factorial:

Letters	Permutations
1	$1=1$
2	$2\times 1=2$
3	$3\times 2\times 1=6$
4	$4\times 3\times 2\times 1=24$
5	$5!=120$
6	$6!=720$
7	$7!=5040$
8	$8!=40320$
9	$9!=362880$
10	$10!=3628800$

In general, $n$ distinct letters admit $n!=n(n-1)\cdots 2\cdot 1$ orderings. Each additional letter multiplies the count by the new total. See factorials. (source: chapter-1.2.11)

Repeated Letters (articles 357–358)

When letters repeat, identical orderings collapse. The rule: divide the all-distinct count by the factorial of each group of repeated letters.

For the word $aaabbc$ (6 letters: $a$ three times, $b$ twice, $c$ once):

$$\frac{6!}{3!\cdot 2!\cdot 1!}=\frac{720}{6\cdot 2\cdot 1}=60$$

(source: chapter-1.2.11)

Proof of the Binomial Coefficient Formula (article 359)

A term $a^{n-k}{b}^k$ in $(a+b{)}^n$ consists of $n$ letters with $a$ repeated $n-k$ times and $b$ repeated $k$ times. Its coefficient is therefore

$$\left(\genfrac{}{}{0ex}{}{n}{k}\right)=\frac{n!}{k!(n-k)!}$$

Euler verifies for the 7th power term by term:

• $a^{6}b$: $\frac{7!}{6!1!}=7$
• $a^5{b}^2$: $\frac{7!}{5!2!}=\frac{7\cdot 6}{1\cdot 2}=21$
• $a^4{b}^3$: $\frac{7!}{4!3!}=\frac{7\cdot 6\cdot 5}{1\cdot 2\cdot 3}=35$

This confirms the descending-fraction rule of ch1.2.10-higher-powers-compound. (source: chapter-1.2.11)

Extension to Multinomials (article 360)

The same counting argument applies when the root has more than two terms. For $(a+b+c{)}^3$, every term is a 3-letter word over the alphabet $\{a,b,c\}$, and its coefficient is the number of distinct orderings of those letters:

$$(a+b+c{)}^3=a^3+3a^{2}b+3a^{2}c+3a{b}^2+6abc+3a{c}^2+b^3+3b^{2}c+3b{c}^2+c^3$$

Setting $a=b=c=1$ gives $27=3^3$ ✓; setting $a=b=1,c=-1$ gives $1^3=1$ ✓. (source: chapter-1.2.11)

Related pages

• factorials
• binomial-theorem
• pascal-triangle
• ch1.2.10-higher-powers-compound
• ch1.2.12-irrational-powers-infinite-series
• algebraic-identities
• powers-and-exponents