How many 5 digit numbers contain all the digits 1,2,3,4,5 and have the property that each pair of adjacent digits has a difference of at least 2? 
How many 5 digit numbers contain all the digits 1,2,3,4,5 and have the property that the difference between each pair of adjacent digits is at least 2?

$-$Question 22, Junior Division AMC 2016

I know that I could do this the long way, but I'm looking for a way I could solve this. Possible examples are 53142, however is there a way to find the answer easily and effectively
P.S. Please use methods familiar to Year 7,8&9  
 A: To quickly find all possibilities, consider $5$ cases:
$1. \ 3\, \_\, \_\, \_\, \_$
$2. \ \_\, 3\, \_\, \_\, \_$
$3. \ \_\, \_\, 3\, \_\, \_$
$4. \ \_\, \_\, \_\, 3\, \_$
$5. \ \_\, \_\, \_\, \_\, 3$
Explanation: $3$ is very convenient since only $1$ and $5$ can be next to it. Also, note that if we apply $x\mapsto 6 - x$ to each numeral, we will get a new solution from the old one; and that is why we like $3$, it is fixed under this map. So, why do we consider this? Because, for each of the above cases, there will always be two subcases where $1$ and $5$ switch places. This is exactly what $x\mapsto 6-x$ does.
Thus, our cases become
$1'. \ 3\, 1\, \_\, \_\, \_$
$2'. \ 5\, 3\, 1\, \_\, \_$
$3'. \ \_\, 5\, 3\, 1\, \_$
$4'. \ \_\, \_\, 1\, 3\, 5$
$5'. \ \_\, \_\, \_\, 1\, 3$
but we can ignore numbers $4.$ and $5.$ since they are just mirror images of $1.$ and $2.$ If we denote with $n_i$ the number of possibilities in $i$-th case, we get total number of $$n = n_1+n_2+n_3+n_4+n_5 = 2n_1 + 2n_2 + n_3 = 4n_1'+4n_2' + 2n_3'.$$
Now, to counting:
$1.'$ Since $4$ and $5$ cannot be next to each other, we must have $2$ sitting on the $4$-th place, i.e. $3\, 1\, \_\, 2\, \_$ and we have two possibilities from here: $3\, 1\, 4\, 2\, 5$ and $3\, 1\, 5\, 2\, 4$. Thus $n_1' = 2$.
$2.'$ and $3.'$ Obviously, there is only one possibility in each case: $5\, 3\, 1\, 4\, 2$ and $2\, 5\, 3\, 1\, 4$. Thus $n_2'=n_3' = 1$.
Finally, $n = 4\cdot 2+4\cdot 1+2\cdot 1 = 14$.

Addendum: You can just ignore the stuff with $x\mapsto 6-x$ and simply write it down. However, if you had similar problem with $2k+1$ numerals, it would halve your work.
A: There are $$5!=120$$ permutations of the numbers 1,2,3,4,5 - so we know that the answer will be less than this number.
Let's list viable solutions which begin with the number 1:


*

*1,3,5,2,4

*1,4,2,5,3


Notice that the conditions you have stated, constrain the sequence in both forward and backward directions.  Also notice that the second sequence is the reverse of the first, starting from another position.
You can start on any of the numbers in the top answer, and place anything which comes before your starting location, after the end.


*

*1,3,5,2,4

*3,5,2,4,1

*5,2,4,1,3

*2,4,1,3,5

*4,1,3,5,2
...


You can select 5 sequences by simply changing the start location.
You also have the option to read the sequences in reverse as the numbers at the end points are two apart.
Therefore, $5\times 2 = 10$ sequences.
There are also some sequences which cannot be permuted but can be reversed because the numbers at either end are less than 2 apart.  These occur because the largest and smallest permissible values are in the centre of the sequence.


*

*2, 4, 1, 5, 3

*3, 1, 5, 2, 4


add these 4 to your previous 10. and you'll have found all 14 


*

*1:    1, 3, 5, 2, 4

*2:    1, 4, 2, 5, 3

*3:    2, 4, 1, 3, 5

*4:    2, 4, 1, 5, 3

*5:    2, 5, 3, 1, 4

*6:    3, 1, 4, 2, 5

*7:    3, 1, 5, 2, 4

*8:    3, 5, 1, 4, 2

*9:    3, 5, 2, 4, 1

*10:   4, 1, 3, 5, 2

*11:   4, 2, 5, 1, 3

*12:   4, 2, 5, 3, 1

*13:   5, 2, 4, 1, 3

*14:   5, 3, 1, 4, 2

A: Take case by case for which is middle number.  Note that in all cases all numbers excluding middle number will have $2$ pairs of consecutive numbers e.g. $1,3,4,5$ has pairs $3,4$ and $4,5$. That being the case, once have one pair consecutive only one possible positioning for remaining pair so that none other consecutive e.g. for $3,4,5,1$ and $4,5$ as consecutive pair must be $1,4,5,3$.
$1$ or $5$ for middle number:  $2$ adjacent numbers must must be $3,4$ or $4,5$;  $3,2$ or $2,1$ respectively else if were e.g. $3,5$ then $3$ adjacent to both $2$ and $4$ and won't work.  $2$ pairs of numbers each with $2$ orderings is $2\times 2=4$ possible numbers for each of $1$ and $5$ as middle number.
$2$ or $4$ for middle number:  $2$ adjacent numbers must be $4,5$ or $1,2$ respectively. So $2$ possible numbers for each.
$3$ as middle number: $2$ adjacent numbers must be $1,5$. Then only one place for $2,4$ to go once they are set.
So $4+4+2+2+2=14$ possible numbers.
