1. In a locality, there are ten houses in a row. On a particular night a thief planned to steal from three houses of the locality. In how many ways can he plan such that no two of them are next to each other?
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By: anil on 05 May 2019 02.37 am
Total number of houses = 10 If first house is robbed, then II is not and if II house is robbed, then III is not and so on. Thus 2 adjacent houses can never be chosen So, number of ways in which three houses can be robbed such that no two of them are next to each other. = $$C^{10 - 2}_3 = C^8_3$$ = $$frac{8 imes 7 imes 6}{1 imes 2 imes 3}$$ = $$56$$
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