piling of two diffrent types of coins and permutation You have an unlimited supply of 2 types of coins(coin-1 and coin-2). You are asked to build a pile of 1000 coins such that the number of coins between any two coin-2 is not equal to five. So what is the maximum number of coin-2 that can be included in the pile?
i know permutation and combination bit well but this question seems different and unique to me, how to tackle such problem? please help, thank you.
 A: Evaluate the pile from bottom to top and list all the locations of coin-2: $\{x_{1},…,x_{m}\}$.
Say that there are no coin-2 pair with five coins between them. Then there are no $i<j$ that satisfy $x_{i}+6=x_{j}$ because otherwise there will be five coins between coin number $i$ and coin number $j$. Consequently, $\{x_{1},…,x_{m}\}$ and $\{x_{1}+6,…,x_{m}+6\}$ are $2m$ distinct integers between $1$ and $1006$. This give us an upper bound for $m$; $2m\leq1006\to m\leq 503$
However, $503$ is not the answer. Notice that $\{x_{1},…,x_{m}\}$ and $\{x_{1}+6,…,x_{m}+6\}$ are identical in modulo $6$ so if we write these in modulo $6$ we will get even number of each of $0,1,2,3,4,5$. Write $1$ to $1006$ in modulo $6$ and you’ll see that $1,2,3,4$ appear $168$ times while $5,0$ appear $167$ times. Means at least one number with $0$ in modulo $6$ and one number with $5$ in modulo $6$ will not appear. Therefore the upper bound need to be revised: $2m\leq 1006-2\to m\leq 502$
Turn out, stacking $6$ coin-2 and $6$ coin-1 alternatingly gives us the desired pile and a total of $502$ coin-2.
