In A Locality, There Are Ten Houses In A Row. On A Particular Night GMAT Problem Solving

Question: 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?

1. 56
2. 73
3. 80
4. 120
5. 224

“In a locality, there are ten houses in a row. On a particular night”- is a topic of the GMAT Quantitative reasoning section of the GMAT exam. The GMAT quant section is mainly prepared to test the calculative skills of the candidates. The candidates must calculate the quantitative sums based on mathematical concepts. The candidates have a concrete concept of mathematics in order to crack GMAT Problem Solving questions. The mathematical problems of the GMAT Quant topic in the problem-solving part can be solved by the candidates’ logical thinking and mathematical ability.

Solution and Explanation:

Approach Solution 1:

The problem statement informs that:

Given:

• There are ten houses in a row.
• A thief steals from three houses in the locality

Find Out:

• The number of ways the thief can plan such that no houses are next to each other.

The total number of choosing 3 house out of 10 houses = C(10,3) = 10!/(7!3!) = 120
The number of ways that a thief can plan such that all three houses chosen is adjacent
= 10 - 3 +1 = 8

The number of ways that a thief can plan such that two houses are adjacent while the 3rd house is not adjacent = 7+7+6+6+6+6+6+6+6 =56 (7 when two adjacent houses lie on the extreme end of the row and 6 in the rest of the cases)

The number of ways that a thief can plan such that all three houses are not adjacent
= 120 - 8 - 56 = 120 - 64 = 56.

Correct Answer: (A)

Approach Solution 2:

The problem statement suggests that:

Given:

• There are ten houses in a row.
• A thief steals from three houses in the locality

Find Out:

• The number of ways the thief can plan such that no houses are adjacent to one another.

Here, as per the given question, the thief has to select 3 houses out of 10 houses.
If it happens without any restriction, it would be 10C3

The question states that houses need to be selected in such a way that no 2 houses are next to each other.
If the first house is chosen, then the thief cannot select the second house. The thief can select the 3rd, 4th or 5th. Hence, for any houses selected by the thief, there exists a restriction of choosing the third and the last house.
Therefore, the thief needs to exclude two houses out of the ten houses in any case.
Thus, the selection of thief becomes (10 - 2)C3 = 8C3 = 56.

Hence, the number of ways the thief can plan such that no houses are adjacent to one another is 56.

Correct Answer: (A)

Approach Solution 3:

The problem statement indicates that:

Given:

• There are ten houses in a row.
• A thief steals from three houses in the locality

The question asks to find out the number of ways the thief can plan such that no houses are adjacent to one another.
If there exist no restrictions, the thief can select the houses in 10C3 ways.

That is we can say, 10C3 = (10 x 9 x 8)/(3 x 2) = 10 x 3 x 4 = 120 ways
Therefore, the thief can have 120 ways to select any 3 houses of the 10 houses.

let’s mark the 10 houses as A, B, C, D, E, F, G, H, I, and J.
If all three houses are set together then we get, ABC, BCD, CDE, DEF, EFG, FGH, GHI and HIJ - as an aggregate of 8 ways.
If two houses are set together and if the two houses are AB, then we get ABD, ABE, ABF, ABG, ABH, ABI and ABJ - as an aggregate of 7 ways.
If the two houses are BC, then we get BCE, BCF, BCG, BCH, BCI and BCJ -as an aggregate of 6 ways.
If two houses are CD, DE, EF, FG, GH, and HI, all these pairs have an aggregate of 6 ways (like BC) since the thief cannot select a house next to each other.
If two houses are IJ, then we get an aggregate of 7 ways ( like AB) since the thief cannot select a house just like the thief can’t select house C for AB.
Therefore, if two houses are set together, there are 2 x 7 + 7 x 6 = 14 + 42 = 56 ways.
Thus, the number of ways the thief can plan such that no houses are adjacent to one another =120 - 8 - 56 = 56 ways

Correct Answer: (A)

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