Aptitude Discussion

Q. |
Two squares are chosen at random on a chessboard. What is the probability that they have a side in common? |

✔ A. |
1/18 |

✖ B. |
64/4032 |

✖ C. |
63/64 |

✖ D. |
1/9 |

**Solution:**

Option(**A**) is correct

Sample space will be total ways of selecting 2 squares (order doesn't matter) out of available 64 squares.

So sample space\(=^{64}C_2\)

Now there are 7 unique adjacent square sets in each row and each column.

i.e. favourable cases will be 7×(8 rows + 8 columns) = 112.

Hence Required Probability

\(=\left(\dfrac{\text{Favourable cases}}{\text{Sample Space}}\right)\)

\(=\dfrac{112}{^{64}C_2}\)

\(=\dfrac{1}{18}\)

Also, **there could be other solution too** if we consider ALL possible squares and not only the smallest squares. (i.e. if we consider that total squares are more than 64).

In that case, let us first calculate the sample space i.e. total number of squares on a chess board.

1, 8x8 square

4, 7x7 squares

9, 6x6 squares

16, 5x5 squares

25, 4x4 squares

36, 3x3 squares

49, 2x2 squares

64, 1x1 squares

Therefore, there are actually:

\(=64+49+36+25+16+9+4+1\)

\(=8^2+7^2+6^2+...1^2=204 \)

squares on a chessboard!

So sample space\(=^{204}C_4\)

Now if we assume that 2 squares can have a side common only if they are of the same dimensions:

Let the length of the smallest square be 1 unit.

So the possibility of selecting 2 squares with a common side (As calculated in the first method above):

having the side length greater than 4 unit length = 0

having the side length of 4 unit = 1×(2 rows + 2 columns) = 4

having the side length of 3 unit = 3×(6 rows + 6 columns) = 36

having the side length of 2 unit = 5×(7 rows + 7 columns) = 70

having the side length of 1 unit = 7×(8 rows + 8 columns) = 112

So total number of favourable cases:

$=4+36+70+112=222$

Hence Required Probability

\(=\left(\dfrac{\text{Favourable cases}}{\text{Sample Space}}\right)\)

\(=\dfrac{222}{^{204}C_2}\)

**Edit:** For an alternative solution, check comment from **Bhavya Shah.**

**Edit 2:** For yet another alternative solution, check comment by **RandomMathMan.**

**B Gogoi**

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**Gaurav Karnani**

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Total number of Cases for a chessboard$ 7*(7*2+1) $ that equals $112$

P= $112/64C2$

Forgot to add seven to the number of cases for the last row. please amend it accordingly.

**RandomMathMan**

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There's a much easier way to do this. On a chessboard, there are 36 squares that border 4 other squares, 24 squares that border 3 other squares, and 4 squares that border only 2 other squares.

$\dfrac{36}{64} \times \dfrac{4}{63} + \dfrac{24}{64} \times \dfrac{3}{63} + \dfrac{4}{64} \times \dfrac{2}{63} = \dfrac{1}{18}$

Exactly how I too solved this. Much simpler than the other alternatives given here.

**Bhavya Shah**

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For each row, there are 7 ways to select adjacent squares and there are 8 such rows.

Similarly for columns, 7 squares could be selected from one column and there are 8 such columns.

So no. of ways of selecting adjacent squares $= 7*8*2$

Total number of ways of selecting squares of chessboard $= ^{64}C_2$

Probability $=\dfrac{7*8*2}{^{64}C_2}=\dfrac{1}{18}$

**Poonam Pipaliya**

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sir,,how 7 squares in each row and column remain..??..!! and how it can be favorable case..??..!!

Lets talk of the rows first. Once you pick up a square, you leave that row, so you have 7 remaining rows.

Similar argument is valid for columns.

I guess the alternative method given in the solution is easier to understand, why don't you have a look at that.

**Arshal**

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Guys, I think I'm right but, as I have been hitting my head on the wall of probability for last 7 hours I'm kinda in concussion...

So is choosing 2 consecutive elements out of 8 equal to choosing 1 element out of 7

**Sanjay Yadav**

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$C(4,1)*2+C(24,1)*3+C(36,1)/C(64,2)$

There are 4 corner squares for which there are only 2 choices for selecting next square....

Similarly, 24 squares for which there are 3 choices and remaining 36 have 4 choices

So, the answer should be 1/9

Correct me if i am wrong!!!

Answer will be 1/18....

As I counted every square non repeated

So, total no of unique square=

$\dfrac{((C(4,1)*2+C(24,1)*3+C(36,1)*4)/2)}{C(64,2)}$

$=\dfrac{1}{18}$

**Sanjay Yadav**

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C(4,1)*2+C(24,1)*3+C(36,1)/C(64,2)

there are 4 corner squares for which there are only 2 choices for selecting next square....

similarly 24 squares for which there are 3 choices and remaining 36 have 4 choices

so the answer should be 1/9

correct me if i am wrong!!!

My method:

Sides of squares on the boundary of the board are not shared.

All smallest sides inside the board are shared.

Total no. of sides =8x9x2(Both directions, x and y or vertical and horizontal)=144

Unshared sides=8x4(all 4 boundaries)=32

No. of shared sides=144-32=112

Total no. of possible pairs of squares=2016(as calculated before)

So, the probability=112/2016=1/18

Hope it is OK.