3

Given $2n$ letters, two of each of $n$ types. How many arrangements of these letters are there with no pair of consecutive letters the same? Please check my solution and tell me where my mistake is!

So such arrangements where the pairs of identical letters are distinguishable. Then $∑^n_k (−1)^k2^k(2n−k)! C(n,k)$ gives the number of such arrangements by inclusion-exclusion (there are $(2n−k)!$ ways for $k$ given pairs to be together (by merging the those pairs into single elements) and the rest arbitrarily ordered, $2^k$ ways to order within those given pairs, and $C(n, k)$ ways to pick those $k$ pairs).So we can obtain the formula :$∑^n_k(−1)^k2^k(2n−k)!C(n,k)$.

Johnathan
  • 323
  • What makes you think there is an mistake? –  Mar 26 '17 at 00:40
  • my professor said thre is a minor mistake but told me to figure out myself :( @ByronSchmuland – Johnathan Mar 26 '17 at 13:05
  • You should specify the range of $k$ in the sum, but apart from that it looks OK to me. –  Mar 26 '17 at 13:07
  • http://math.stackexchange.com/questions/580435/number-of-2n-letter-words-using-double-n-letter-alphabet-without-consecutiv –  Mar 26 '17 at 20:06

1 Answers1

2

Calculating the number $a_n$ of valid arrangements of $2n$ letters with two of each of $n$ types for small values $n=1,2$ we obtain $\color{blue}{a_1=0}$ and $\color{blue}{a_2=2}$. So, there are no valid arrangements of length two and two valid arrangements of length four. Assuming an alphabet $\{a,b\}$ we obtain \begin{align*} \color{blue}{abab}\qquad\text{and}\qquad \color{blue}{baba} \end{align*}

Checking the formula \begin{align*} b_n=\sum^n_k (-1)^k2^k(2n-k)! C(n,k) \end{align*} for small values $n=1,2$ results in $b_1=0$ and $b_2=8$, the second value being false.

But in fact there is only a small correction necessary. When considering $2n$ letters with two of each of $n$ types, the number of different arrangements is not $(2n)!$, but instead \begin{align*} \frac{(2n)!}{2^n} \end{align*} For each occurrence of a letter of a specific type two arrangements are to identify (when exchanging the letters of the same type) giving a total of $2^n$ arrangements to identify.

We conclude, the correct formula is \begin{align*} a_n=\frac{b_n}{2^n}=\sum^n_k (-1)^k2^{k-n}(2n-k)! C(n,k) \end{align*} starting with \begin{align*} 0,2,30,264,39480,2631600,241133760,\ldots \end{align*}

Markus Scheuer
  • 108,315