In a locality there are ten houses in a row. On a particular night the

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?

A. 56
B. 73
C. 80
D. 120
E. 224

prabsahi · Accepted Answer

So we have to select 3 houses out of 10.If we were to do this without any restriction it would have been 10C3

Now ,If the first house is selected then the second is not selected and then we can select 3rd or 4th or 5th or..but whicever we select again there is a restriction while selected 
the third or last house. So in short ,we will be excluding two houses out of 10 in any cases.

So our selection becomes (10-2) C 3 that is 8C3 =56

Press Kudos if it helps!!

Princ · Answer

OA:A

Total number of selecting  3  house out of  10= C(10,3) = \frac{10!}{7!3!}=120 
 
Number of ways such that all 3 houses selected are adjacent  = 10-3+1=8

Number of ways such that 2 houses are adjacent while 3rd house is not adjacent =  7+7 + 6+6+6+6+6+6+6   =56  ( 7  when 2 adjacent houses are on extreme end of row and  6  in remaining cases)

Number of ways that all 3 houses are not adjacent  = 120-8-56=120-64=56

Abhi077 · Answer

ScottTargetTestPrep  waiting for your solution on this.

T1101 · Answer

Very nice explanation! But I still do not understand why the method of  \frac{10*8*6}{3!}  is incorrect in this case... I am gettin confused on many of these questions...

ScottTargetTestPrep · Answer

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

prabsahi · Answer

I hope you mean 10*9*8/3! 
 
Please tell me what you line of thought ?

bumpbot · Answer

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In a locality there are ten houses in a row. On a particular night the

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