John computes the sum of the elements of each of the 15 two-element subsets of $\{1,2,3,4,5,6\}$. What is the sum of these 15 sums?

It might be interesting to see if we can discover a pattern here.

Note

Three elements....two at  time  (1,2) (1,3) (2,3)....sum  =   12

Four elements .... two at a time  (1,2) (1,3) (1,4) (2,3) (2,4) (3,4)  .....sum  =  30

Five elements....two at a time  (1,2) (1,3) (1,4) (1,5) (2,3) (2,4) (2,5) (3,4) (3,5) (4,5)  ....sum  = 60

Note that for N elements taken  2 at a time......the pattern of sums seems to be

(N - 1) ( N ) ( N+ 1)  / 2  

So....for 6 elements taken 2 at a time.....the sum should be

(5) (6) (7)  / 2  =     210 / 2  =   105

Verify for yourself that this is true.......!!!!!!

As an alternative answer  to this.....

Note that each  element  will be summed  5  times

So we have  the sum  (1 + 2 + 3 + 4 + 5 + 6)   summed 5 times

And the sum of  the frist  6 elements  =  (6) (7)  / 2  =  21 

So......the total sum will be  

(5) * [ (6) * ( 7)  / 2 ]   =    (N -1) (N) ( N + 1)  / 2     which is the correct "formula"

John computes the sum of the elements of each of the 15 two-element subsets of
$\{1,2,3,4,5,6\}$.
What is the sum of these 15 sums?

15 two-element subsets:

$\begin{array}{|llllll|} \hline &1,& 2,& 3,& 4,& 5,& 6 \\ \hline &1,2 & 2,3 & 3,4 & 4,5 & 5,6 \\ &1,3 & 2,4 & 3,5 & 4,6 & \\ &1,4 & 2,5 & 3,6 & \\ &1,5 & 2,6 & \\ &1,6 \\ \hline \end{array}$

$\begin{array}{|lcll|} \hline \text{the numbers are: } && 5\times 1 \\ &+& 5\times 2 \\ &+& 5\times 3 \\ &+& 5\times 4 \\ &+& 5\times 5 \\ &+& 5\times 6 \\ \hline &=& 5\times (1+2+3+4+5+6) \\ &=& 5\times \left(\frac{1+6}{2}\right)\times 6 \\ &=& 5\times \left(\frac{7}{2}\right)\times 6 \\ &=& 105 \\\\ &=& (n-1)\times \left(\frac{1+n}{2}\right)\times n \\ &=& 3\times \binom{n+1}{3} \\ \hline \end{array}$