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
A
`120`
B
`240`
C
`360`
D
`720`
Text Solution
Verified by Experts
The correct Answer is:
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`
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CENGAGE-PROGRESSION AND SERIES-ARCHIVES (MATRIX MATCH TYPE )