How many $n-$digit number that contain only digits $ 1,2,3,4,5,6$ How many $n- $digit numbers can be formed from the digits $1,2,3,4,5$ and 6, which contains the  numbers $1$ and $2$ as neighbours.
Let $p_n$ be the number of n-digit numbers which consist only of the digits 1,2,3,4,5,6, which contains the  numbers $1$ and $2$ as neighbours.
$$p_{n+1}=p_{n}+(?)$$
Now How to find  $(?)$ 

Start by making $n$-symbol words from the five symbols $[12],3,4,5,6$; there are $\binom{n+4}{4}$ ways to do this (use "stars and bars".)  For each such word, you have two possible orderings of the $1$ and $2$.  So
$$p_n=2\binom{n+4}{4}.$$
And $p_{n+1}-p_n=2\binom{n+5}{4}-2\binom{n+4}{4}=2\binom{n+4}{3}$.
