Triple summation having $i,j,k$ as variables 
Finding value of


$(1)\ \displaystyle \mathop{\sum\sum\sum}_{1\leq i<j<k\leq n}ijk$


$(2)\ \displaystyle \mathop{\sum\sum\sum}_{1\leq i\leq j\leq k\leq n}ijk$

For $(1) $ what I have done as
For without any restriction as
$ (S)=\displaystyle \mathop{\sum\sum\sum}_{1\leq i,j,k\leq n}ijk=\bigg(\sum^n_{i=1}i\bigg)^3=\bigg(\frac{n(n+1)}{2}\bigg)^3$
For exactly one equal
$\displaystyle S'=\displaystyle \mathop{\sum\sum\sum}_{1\leq i=j,k\leq n}ijk=\bigg(\sum^n_{j=1}j^2\bigg)\bigg(\sum^n_{k=1}k\bigg)=\bigg(\frac{n(n+1)(2n+1)}{6}\bigg)\bigg(\frac{n(n+1)}{2}\bigg)$
There are 3 such pairs $( i=j,j=k,k=i)$
All three equal
$S''=\displaystyle \mathop{\sum\sum\sum}_{1\leq i=j=k\leq n}ijk=\sum^n_{k=1}k^3=\bigg(\frac{n(n+1)}{2}\bigg)^2$
Now we have
$\displaystyle \mathop{\sum\sum\sum}_{1\leq i<j<k\leq n}ijk=S-3S'+S''$
$\displaystyle =\frac{n^2(n+1)^2}{4}\bigg(\frac{3n^2-n+4}{6}\bigg)$
But this is not correct, please tell me where I am wrong
Also please tell me part $2$

It is better to make each summation explicit.  You can then evaluate them one by one.  You are looking for
$$\sum_{k=2}^n\sum_{j=2}^{k}\sum_{i=1}^{j-1}ijk$$
where the $2$ comes because we require $i \lt j$.  Now evaluate from the inside out, so this becomes
$$\sum_{k=2}^n\sum_{j=2}^{k}\sum_{i=1}^{j-1}ijk=\sum_{k=2}^n\sum_{j=2}^{k}\frac{j(j-1)}2jk$$
Keep going.  For the second one, you just update the lower limits to $1$ and the upper $i$ limit to $j$.
