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Question 66
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?
Solution
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.
= $$C^{10 - 2}_3 = C^8_3$$
= $$\frac{8 \times 7 \times 6}{1 \times 2 \times 3}$$
= $$56$$
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