# Let A={a,b,c,d} 1. Find all combinatorial lines in A^2. How many combinatorial lines are there? 2. Let n in NN. Prove that the number of combinatorial lines in A^n equals 5^n-4^n

Let $A=\left\{a,b,c,d\right\}$
1. Find all combinatorial lines in ${A}^{2}$. How many combinatorial lines are there?
2. Let n in $\mathbb{N}$. Prove that the number of combinatorial lines in ${A}^{n}$ equals ${5}^{n}-{4}^{n}$
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1) For any $l\in A$ , let us define
${B}_{1}l=\left\{\left(l,x\right):x\in A\right\}$
${B}_{2}l=\left\{\left(x,l\right):x\in A\right\}$
$C=\left\{\left(x,x\right):x\in A\right\}$
These above mentioned are the three combinatorial lines in ${A}^{2}$. There is only one combinatorial line of type C , which is C itself. |A| = 4, there are four combinatorial lines ${B}_{1}$, ,t in A and there are four combinatorial lines in ${A}^{2}$.These are all the combinatorial lines in ${A}^{2}$. Hence there are a total of 9 combinatorial lines in ${A}^{2}$.
2) Consider any Subset $E\subseteq \left\{1,\dots .n\right\},E\ne \mathrm{\varnothing }$ . For every , fix any ${a}_{i}$ in A.
Then we get a combinatorial line defined as:
${B}_{E,\left\{ai\right\}Ii\in E}=\left\{{x}_{1},\dots \dots ..{x}_{n}\right\}\in An:\left({x}_{i}={a}_{I\mathrm{\forall }i\in E}\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\left({x}_{i}={x}_{i\mathrm{\forall }i\in E}\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\left(x\in A\right)\right\}$
We further note that reach combinatorial line is of the form , since there is only one free coordinate in a combinatorial line.
Now , for any $F\in \left\{1,\dots n\right\}$ , the number of subsets of $\left\{1\dots .n\right\}$ of size F is equal to $\left(\frac{n}{k}\right)$
Further , for any subset $E\subseteq \left\{1,\dots .n\right\},|E|=F$ , the number of choices for the remaining coordinates $i\in \left\{1,2\dots .n\right\}$
$i\notin E$ is equal to ${|A|}^{n}-|E|={|A|}^{n-F}={4}^{n-F}$
Hence the total number of combinatorial is equal to :
$\sum _{F=1}^{n}=\left(\sum _{F=0}^{n}\left(\frac{n}{k}\right){4}^{n-F}\right)-{4}^{n}$
$=\left(\sum _{F=0}^{n}\left(\frac{n}{k}\right){4}^{-F}\right)-{4}^{n}$
$={\left(4+1\right)}^{n}-{4}^{n}$
$={5}^{n}-{4}^{n}$
By Binomial Theroem .