The sum `sum_(k=1)^(10)underset(i lt j lt k)underset(j=1)(sum^(10))sum_(i=1)^(10)1` is equal to

Correct Answer - A
`(a)` `sum_(k=1)^(10)underset(i lt j lt k)(sum_(j=1)^(10))sum_(i=1)^(10)1`
`=(1)/(6)sum_(k=1)^(10)underset(i ne j ne k)(sum_(j=1)^(10))sum_(i=1)^(10)1`
As in `sum_(k=1)^(10)underset(i lt j lt k)(sum_(j=1)^(10))sum_(i=1)^(10)1` , we have sum of terms for `i lt j lt k`,` i lt k lt j`,
`j lt i lt k`, `j lt i lt k`, `k lt i lt j`, ` k lt j lt i` and sum for each inequality is same
`:.sum_(k=1)^(10)underset(i lt j lt k)(sum_(j=1)^(10))sum_(i=1)^(10)1`
`=(1)/(6)sum_(k=1)^(10)underset(i ne j ne k)(sum_(j=1)^(10))sum_(i=1)^(10)1`
`=(720)/(6)`
`=120`