adilsafder wrote:
In a locality, there are ten houses in a row. On a particular night thief planned to steal from three houses of the locality. In how many ways can he plan such that no two of them next to each other?
A. 56
B. 73
C. 80
D. 120
E. 224
If there are no restrictions, the thief has 10C3 = (10 x 9 x 8)/(3 x 2) = 10 x 3 x 4 = 120 ways to choose any 3 of the 10 houses.
Now, let’s label the 10 houses are A, B, C, D, E, F, G, H, I, and J. If all three houses are together, we have ABC, BCD, CDE, DEF, EFG, FGH, GHI and HIJ - a total of 8 ways.
If exactly two houses are together and the two houses are AB, we have: ABD, ABE, ABF, ABG, ABH, ABI and ABJ - a total of 7 ways. If the two houses are BC, we have BCE, BCF, BCG, BCH, BCI and BCJ - a total of 6 ways. If the two houses are CD, DE, EF, FG, GH, and HI, each of these pairs have a total of 6 ways (same as BC) since we can’t choose a house immediately to their right or to their left. If the two houses are IJ, then they have a total of 7 ways (same as AB) since we can’t choose house just like we can’t choose house C for AB. Therefore, if exactly two houses are together, there are 2 x 7 + 7 x 6 = 14 + 42 = 56 ways.
So there are 120 - 8 - 56 = 56 ways to choose the 3 houses without any two of them being next to each other.
Answer: A
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