Sum Subsets

I just iterated the values like this:

$\{1,2,3\} \{1,3,4\},\{1,4,5\} ...\space... = 48 \space sets$ . $\{2,4,6\} \{2,5,7\},\{2,6,8\} ...\space... = 46 \space sets$ . $\{3,4,7\} \{3,5,8\},\{3,6,9\} ... \space...= 44 \space sets$ . $...\space...\space...\space...\space...\space...\space$ $...\space...\space...\space...\space...\space...\space$ $\{23,24,47\} \{23,25,48\},\{23,26,49\} ...\space... = 4 \space sets$ . $\{24,25,49\}\{24,26,50\}=2\space sets$ Now $2+4+...\space..\space...+44+46+48 = 600$ $\boxed{600}$

The smallest three element subset would be ${1,2,3}$ The smallest number that can be represented as the sum of the other two is $3$ For an integer $n$ it can be expressed as the sum of two different positive integers the following ways $(n-1)+1$ , $(n-2)+2$ , $...$ until $\lfloor \frac{n}{2}\rfloor + (\lfloor \frac{n}{2}\rfloor)+1$ (if n is odd) and $(\lfloor \frac{n}{2}\rfloor)-1 + (\lfloor \frac{n}{2}\rfloor)+1$ (if n is even). So, the number of ways $n$ can be expressed as the sum of two different positive integers is to take the number of positive integers less than $n$ and divide it by 2, that is $\lfloor\frac{n-1}{2}\rfloor$ The we have the number of ways $n$ can be express as the sum of two different positive integers as $3,4=1$ , $5,6=2$ ... $49,50=24$ which gives us $1+1+2+2+...+24+24=2(\frac{24*25}{2})=24*25=600$ (P.S I'm quite new to using LaTeX so the formatting might be a little bad. My explanation might not be that clear too. Forgive me for my long explanation)

Let each subset be of type (A , b,c ) where A =b+c . Now ,clearly A cannot be equal to 2 or 1 since neither of A,b,c are same. Thus, A= 3 or 4 or ... or 50. If A is an odd number , then the number of possible subsets for that particular value of A are $\frac{A-1}{2}$ and if A is even, the it is $( \frac{A}{2} -1)$ .This can be easily seen by considering the examples of A=3 ,A=4 ,A=5,A=5 (from induction). It is noteworthy that (3,2,1) is same as (3,1,2).

let me 'TRY' to explain this.

1) we have to choose 3 numbers. consider the first 2 numbers should sum up to the 3rd number. note that these 3 solutions placed in any order will make only 1 set.

2) if the 3rd number is x then the first two can be (x-1 , 1), (x-2 , 2) ...... (1, x-1). therefore if x is placed as the 3rd number it can have x-1 combinations in the 1st two places.

therefore 49 can have 48 solutions , 48 can have 47 solutions, ....., 1 will have 0 solutions. but for even numbers one of the solutions will be (x/2 , x/2). but we cant take 2 same numbers as they are not available in the set .

therefore an even number x will have (x-2) solutions and an odd number y will have (y-1)

3) but as you can see (x-1 , 1) and (1 , x-1) are the same solution just in different orders, this will happen with every such pair.

Therefore for an even number x there will be (x-2)/2 solutions and for a odd number y there will be (y-1)/2 solutions

4) therefore 49, 48, 47, 46, ..... , 1 will have 24, 24, 23, 23,.......0 solutions respectively.

5) therefore answer is (24+24+23+23+22+22+......+0)

= 2(24+23+22+.....+0) = 600

Let's list out to see pattern

1 can pair with 2 to get 3 {1,2,3}, pair with 3 to get 4 {1,3,4}. The last pair for the 1 is 49 to get 50 {1,49,50}.

We only need to pair 2 elements because the last one is already determined by first two numbers.

Write in array for first two elements in subset.

1:| 2 3 4 ......... 49 ( 48 subsets)

2:| 3 4 5 ......... 48 ( 46 subsets)

3:| 4 5 6 ......... 47 ( 44 subsets)

:

:

24:| 25 26 ( 2 subsets)

The last row would be "24" because if we let "25" as a first element, it cannot be paired with any higher number to have the sum less than 50. We also cannot pair with lower otherwise it will duplicate the subset determined by lower first value.

The number of three-element subset that satisfies this condition is 48+46+44+.......+2 = 50*12 =600

I count manually, but knowing the pattern before count it. {a,b,c} a is the first sub sum b is the second sub sum c is the result c<50 The pattern is {(a+1),(b-2),c} So, if every a+1, the b-2 probability is added. Like this : 1 - 48 2 - 46 3 - 44 4 - 42 5 - 40 6 - 38 7 - 36 8 - 34 9 - 32 10 - 30 11 - 28 12 - 26 13 - 24 14 - 22 15 - 20 16 - 18 17 - 16 18 - 14 19 - 12 20 - 10 21 - 8 22 - 6 23 - 4 24 - 2 So, add all 50 multiplies bt 24/2 is 600