SOLUTION: how many ways there are to remove three elements from the sequence 1 4 2 5 7 3 8 9, so that the remaining from an increasing sequence

Question 968110: how many ways there are to remove three elements from the sequence 1 4 2 5 7 3 8 9, so that the remaining from an increasing sequence

Found 2 solutions by KMST, quixote:
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
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 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 numbers,
along with 2 and 3.
That makes 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 number,
so we can do that different ways.
We found ways
"to remove three elements from the sequence 1 4 2 5 7 3 8 9, so that the remaining from an increasing sequence".

Answer by quixote(1) (Show Source): You can put this solution on YOUR website!
Make a graph of the sequence, it will help to visualize
We need to keep 5 numbers:
Case 1: begin with 1 (
case1A: begin with 14 and choose 3 from (5789)----> 4 ways
case1B1: begin with 12 and choose 3 from (5789)----> 4 ways
case1B2: begin with 123 and then we left with (89)---> 1 way
case1C: begin with 15 and then we left with (789)----> 1 way
Case 2: Begin with 4 and then we left with (5789) -----> 1 way
Case 3: Begin with 2 and then we again left with (5789)----> 1 way
Hence total #ways = 12
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