a + c = 2b
Since they are all integers, 2b is even.

We can start by considering different values of b first.

b cannot be 1 since 2b = 2, and no two distinct integers in the set can add up to 2.
b cannot be 200 since 2b = 400, and no two distinct integers in the set can add up to 400.

So 2 ≤ b ≤ 199

a and c are also odd since we don't have two identical copies of b in the set S


When b = 2, 2b = 4
Possible a and c combinations : 1+3 = 4
So there is only 1 set, ｛1,2,3｝

When b = 3, 2b = 6
Possible a and c combinations : 1 + 5 = 6, 2 + 4 = 6
So there are 2 sets, ｛1,3,5｝ and ｛2,3,4｝

When b = 4, 2b = 8
Possible a and c combinations : 1 + 7 = 8, 2 + 6 = 8, 3 + 5 = 8
So there are 3 sets, ｛1,4,7｝,｛2,4,6｝, ｛3,4,5｝

...
...

When b = 100, 2b = 200
Possible a and c combinations : 1 + 199 = 200, 2 + 198 = 200, 3 + 197 = 200,... 99 + 101 = 200

So there are 99 nice sets.


Notice a pattern : the number of sets for each value of b, is always 1 smaller than b.

If we add up all the b from 2 to 100, the number of nice sets will be the sum from 1 to 99.

1 + 2 + 3 + ... + 97 + 98 + 99

Use the Gaussian pairing method.
1 + 99 = 100
2 + 98 = 100
...
49 + 51 = 100
50 is unpaired but its value is equal to half a pair.

We can say there are 49½ pairs or 99/2 pairs.

This corresponds to the formula for such a Sum, ½n(n+1)

= ½(99)(1 + 99)
= 4950

Likewise,

When b = 199, 2b = 398
Possible a and c combinations : 198 + 200 = 398
So there is only 1 set, ｛198,199,200｝

When b = 198, 2b = 396
Possible a and c combinations : 196 + 200 = 396, 197 + 199 = 396
So there are 2 sets, ｛196,198,200｝ and ｛197,198,199｝

When b = 197, 2b = 394
Possible a and c combinations : 194 + 200 = 394, 195 + 199 = 394, 196 + 198 = 394
So there are 3 sets, ｛194,197,200｝,｛195,197,199｝, ｛196,197,198｝

...
...

When b = 101, 2b = 202
Possible a and c combinations : 2 + 200 = 202, 3 + 199 = 202, 4 + 198 = 202,... 100 + 102 = 202

So there are also 99 nice sets.


This is basically a 'mirror' of b = 2 to b = 100


So the number of nice sets is also 4950


Total = 4950 × 2 = 9900

After 2 cards are removed, 38 cards are left.

Number of ways that 2 cards can be selected from 38 with no restrictions

= 38C2
= 703


For the required case (both cards are a pair),

Number of ways whereby both cards are the remaining two cards of the same number as the ones that were removed = 1

(Eg. If 1,1 were removed, there is only 1 way to pick 1,1)


Number of ways whereby both cards are a different number from the ones that were removed

= 9 × 4C2
= 9 × 6
= 54


(There are 9 possible other different numbers. And for each of them, there are 4 indistinguishable cards so selecting any 2 will suffice.

This is where the 4C2 comes from. 6 ways to pick.

Eg.

We could have (9a,9b) , (9a,9c) , (9a,9d) , (9b,9c), (9b,9d) and (9c, 9d) if we actually tried to distinguish them by assigning a different letter. So, even though they are indistinguishable, the number of ways with an identical outcome is 6.)


Required probability = number of ways that fit the requirement / total number of ways

= (54 + 1) / 703
= 55/703

m = 55, n = 703

m + n = 55 + 703 = 758

A convex polygon has a minimum of 3 sides.

Number of ways to choose 3 points out of 10 to form the vertices for a triangle
= 10C3

Number of ways to choose 4 points out of 10 to form the vertices for a convex quadrilateral
= 10C4

Number of ways to choose 5 points out of 10 to form the vertices for a triangle

= 10C5

...
...

Number of ways to choose 10 points out of 10 to form the vertices for a decagon
= 10C10


Total

= 10C3 + 10C4 + 10C5 + ... 10C10
= 968

However, if you have already learnt that the sum of all possible combinations of n distinct objects/items is 2ⁿ,


Then you can just do 2^10 - 10C0 - 10C1 - 10C2

= 1024 - 1 - 10 - 45
= 968