P & C-6 (XAT-2011)

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?

(A)56      (B) 73        (C) 80        (D) 120   (E) None of these

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Solution:

For simplicity first we consider only 5 houses in a row:

√		√		√	1

Number of possibilities = 1 = 3c3

Next we consider 6 houses in a row:

√		√		√
√		√			√	2
√			√		√	1
	√		√		√	1

Number of possibilities = (2+1)+1 = ∑2+∑1 =  4 =  4c3

Next we consider 7 houses in a row:

√		√		√
√		√			√
√		√				√	3
√			√		√
√			√			√	2
√				√		√	1
	√		√		√
	√		√			√	2
	√			√		√	1
		√		√		√	1

Number of possibilities in case of 7 houses = (3+2+1)+(2+1)+1 = ∑3+∑2+∑1 =10 = 5c3

If we consider this logic, the number of possibilities for the thief in case of 10 houses 

= 8c3 = (8*7*6)/(3*2*1) = 56

Answer (A)