Question 968110
1 4 2 5 7 3 8 9 is a sequence of 8 digits (I assume there was no 6).
The digits 1, 2, and 3 are followed by digits that are greater.
The digit 8 and 9 are not followed by any digit with lesser value.
The inversions (deviations from increasing order) happen when
4 is followed by 2,
4 is followed by 3,
5 is followed by 3, and
7 is followed by 3.
So, there are four ordered pairs that make the sequence deviate from the increasing order:
(4,2), (4,3), (5,3), and (7,3).
Removing the set of elements {4,5 7} is {{{red(1)}}} way to eliminate all four inversions.
Can we keep the number 4?
It is involved in 2 inversions: (4,2), and (4,3).
If we want to keep 4, we need to remove at least two numbers: 2 and 3.
If we are going to remove 3 numbers, along with 2 and 3, but we want to keep 4,
we can remove any of the other {{{8-3=5}}} numbers,
along with 2 and 3.
That makes {{{red(5)}}} different sets of 3 numbers.
Of course, if we remove number 4, we eliminate the inversions (4,2), and (4,3),
and we do not need to remove the entire set {4,5 7}.
We can keep 5 and/or 7, as long as we remove number 3,
which would eliminate the other two inversions: (5,3), and (7,3).
How many ways can we chose a set of 3 numbers to remove, including 4 and 3?
Along with 4 and 3, we can remove any of the other {{{8-2=6}}} number,
so we can do that {{{red(6)}}} different ways.
We found {{{red(1)+red(5)+red(6)=highlight(12)}}} ways 
"to remove three elements from the sequence  1 4 2 5 7 3 8 9, so that the remaining from an increasing sequence".