Bunuel wrote:
How many different arrangements of the digits 1, 2, 3, 4, and 5 are possible where no two consecutive digits are adjacent?
A. 42
B. 21
C. 14
D. 10
E. 6
As it is hard to calculate this directly, we'll start by writing down numbers and see how it goes.
This is an Alternative approach.
If we start with a 1 then the next number must be 3,4 or 5.
Starting with 1-3 means that 5 must come next (as 2 and 4 are adjacent to 3) then 2 then 4.
Similarly, 1-4-2-5-3 must follow from 1-4.
However, if we try to continue 1-5-3 we become stuck as both 4 and 2 are impossible.
So, there are 2 numbers starting with 1.
Let's start with 2:
2 can be followed by 4 or 5. Building these gives
2-4-1-3-5 and 2-4-1-5-3
2-5-1- impossible! and 2-5-3-1-4
We have a 3 numbers starting with 2.
Moving on to 3, which can be followed by 1 or 5:
3-1-4-2-5 and 3-1-5-2-4
3-5-2-4-1 and 3-5-1-4-2
4 more options.
Finally, we can list another 3 numbers for 4 and 2 for 5.
In total, we have 2+3+4+3+2=14 options.
(C) is our answer.
** Note: as 1,2,3,4,5 is a symmetrical sequence (5 and 1 are 'end numbers' and 2 and 4 are 'inbetween numbers') then we know that 1 and 5 have the same number of sequences without listing them. Similarly, 2 and 4 have the same number of sequences.