Take a smaller example of this form where among 5 days we need to select 2 non consecutive days.
Instead of choosing 2 days and trying to insert gaps b/w them lets do the opp. This is because in b/w 2 days we might be having more than 1 gap. Then there will be many possibilities to fill spaces b/w 2 days.
We do like, choose the remaining days i.e. 5-2=3. And we try to insert these 2 days in b/w the gaps formed by these 3 days. Note that we can insert only 1 day in each gap else filling a single gap with these 2 days will make them consecutive.
Another imp point to be noted is, we can fill gaps before the 1st day (out of the 3 days) as well as after the last day(out of the 3 days).
Elaboration:
$_ D _ D _ D _$ these are the possible gaps to be filled by our 2 days.
We can fill these 4 gaps with 2 days in C(4,2) ways. How?
I will denote the day inserted as M.
1) $ \underline{M} D \underline{M} D D $
2) $ \underline{M} D D\underline{M} D $
3) $ \underline{M} D D D \underline{M} $
4) $ D\underline{M} D \underline{M} D $
5)$ D\underline{M} D D \underline{M} $
6) $ D D \underline{M} D \underline{M} $
So how to generalize?
Given n no. of days and we need to choose r non-consecutive days, we take the rest of the days i.e. (n-r).
Now no. of gaps that can be filled is (n-r+1).
No. of ways to select r days from (n-r+1) is C(n-r+1,r).
In this sum, n=31 and r=5, so answer becomes C(31-5+1,5)=C(27,5).
Ref: https://math.stackexchange.com/questions/490405/choosing-numbers-without-consecutive-numbers
