If there are 10 houses and the thief needs to choose from 3 houses,

the total ways in which he can choose 3 houses are \(10c3 = \frac{10*9*8}{3*2} = 120\) ways

The condition is that he can't steal from 2 houses that are next to each other.

One way of finding that out is finding how many such combinations are there are reducing that number

from the total combinations possible, which is 120.

To find out how many ways are there if he steals from houses that are next to each other.

If the houses are next to each other, there are 9 ways of choosing the first 2 houses,

and 8 ways of choosing the next house, so a total of 9*8(72) such combinations.

However, there will be 8 combinations which will end up getting counted twice because

_ _ _ _ _ _ _ _ _ _

is the same as

_ _ _ _ _ _ _ _ _ _

That makes a total of 64 combinations(72 - 8) where the houses he steals from are next to each other.

Hence, the total possibilities are 120 - 64 = 56(Option A)

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