Can you solve them?
The math question is fun. Seems like there are two kinds of (i,j)-pairs to look for.
(1) Those that leave only segments that consist of consecutive elements in multiples of 4.
m=1 : (1,2) (1,6) (5,6)
m=2 : all those from before, and also (1,10) (5,10) (9,10)
m=3: all those from before, and also (1,14) (5,14) (9,14) (13, 14)
So the number of ``good'' combos of this sort is
1+2+ ... + m + (m+1) = (m+1)(m+2)/2 = (m^2 + 3m + 2)/m
(2) Those that allow for `skipping sequences'.
m=2 : removing (2,9) allows for arithmetic sequences (1 3 5 7) and (4 6 8 10)
m=3: removing (2,13) allows for arithmetic sequences (1 4 7 10) , (3 6 9 12) and (5 8 11 14)
BUT also, (2,9) still works and so does (6 13), which is just (2 9) shifted over by 4.
The removing pairs of the skipping type are therefore:
m=2 : (2,9)
m=3: (2,9) (6,13) and (2 13)
m=4: (2,9) (6,13) (10,17) and (2 13) (6 17) and (2 17)
1+2+ ... + (m-2) + (m-1) = (m-1)m/2 = (m^2 - m)/m
Overall, the number of ``good'' combos is
(m^2 + 3m + 2)/m + (m^2 - m)/m = m^2 + m +1
Now, the total number of combinations is
nC2 = n! / 2!(n-2)! = (n-1)n/2 = (4m+2)(4m+1)/2 = 8m^2 + 6m+1
Therefore, the probability of drawing a ``good'' pair (i,j) is
Pm = (m^2 + m +1) / (8m^2 + 6m+1) > (m^2 + m +1) / (8m^2 + 8m+8) = 1/8
QED.
I can solve all the questions easily
The math question is fun. Seems like there are two kinds of (i,j)-pairs to look for.
(1) Those that leave only segments that consist of consecutive elements in multiples of 4.
m=1 : (1,2) (1,6) (5,6)
m=2 : all those from before, and also (1,10) (5,10) (9,10)
m=3: all those from before, and also (1,14) (5,14) (9,14) (13, 14)
So the number of ``good'' combos of this sort is
1+2+ ... + m + (m+1) = (m+1)(m+2)/2 = (m^2 + 3m + 2)/m
(2) Those that allow for `skipping sequences'.
m=2 : removing (2,9) allows for arithmetic sequences (1 3 5 7) and (4 6 8 10)
m=3: removing (2,13) allows for arithmetic sequences (1 4 7 10) , (3 6 9 12) and (5 8 11 14)
BUT also, (2,9) still works and so does (6 13), which is just (2 9) shifted over by 4.
The removing pairs of the skipping type are therefore:
m=2 : (2,9)
m=3: (2,9) (6,13) and (2 13)
m=4: (2,9) (6,13) (10,17) and (2 13) (6 17) and (2 17)
So the number of ``good'' combos of this sort is
1+2+ ... + (m-2) + (m-1) = (m-1)m/2 = (m^2 - m)/m
Overall, the number of ``good'' combos is
(m^2 + 3m + 2)/m + (m^2 - m)/m = m^2 + m +1
Now, the total number of combinations is
nC2 = n! / 2!(n-2)! = (n-1)n/2 = (4m+2)(4m+1)/2 = 8m^2 + 6m+1
Therefore, the probability of drawing a ``good'' pair (i,j) is
Pm = (m^2 + m +1) / (8m^2 + 6m+1) > (m^2 + m +1) / (8m^2 + 8m+8) = 1/8
QED.
I can solve all the questions easily