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Four unit squares are chosen at random on a chessboard. What is the probability that three of them are of one colour and fourth is of opposite colour?









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Option(D) is correct

The number of ways of choosing 4 squares from 64 is \(^{64}C_4\)

The possible split ups as far as colour is concerned and the number of ways of selecting the square are listed below:

White = 0, Black = 4, No. of Ways = ${^{32}C_0} × {^{32}C_4}$

White = 1, Black = 3, No. of Ways = ${^{32}C_1} × {^{32}C_3}$

White = 2, Black = 2, No. of Ways = ${^{32}C_2} × {^{32}C_2}$

White = 3, Black = 1, No. of Ways = ${^{32}C_3} × {^{32}C_1}$

White = 4, Black = 0, No. of Ways = ${^{32}C_4} × {^{32}C_0}$

The number of ways in which 3 squares are of one colour and fourth is of opposite colour is: 

$P(A) =2\times {^{32}C_1} \times {^{32}C_3}$

\(=(2\times 32)\dfrac{32\times 31\times 30}{2\times 3}\)

The total number of ways of selecting 4 squares is:

\(= P(B)= {^{64}C_4}\)

\(=\dfrac{64\times 63\times62\times61}{2\times3\times4}\)

The required probability



Edit: For an alternative solution, check comment by Laurence.

(9) Comment(s)


Let the combination, $WWWB$ be a solution of which there can be 4!/3! or 4 permutations. Similarly, there exists another such solution, $BBBW$ 

Now there are 32 White squares and 32 Black squares and the prob. of selecting either sequence is, 


Final answer would, 

$P*4*2 = 0.49960$


Since the question specifically asks that fourth square be the one with the opposite colour, aren't all the answers wrong?

There should only be one permutation: $BBBW$ or $WWWB$. Meaning the actual answer should be $P*2 = 160/1281$.


Actually, there is an easy way.

If we want to have 3 squares having the same color and the other having the opposite color, then these 3 could be either white and the fourth is black(and knowing that a chess consists of 64 squares; 32 white and 32 black) then ${^{32}C_3} \times 32=158720$ or these 3 could be black and the fourth is white. 

So, we will do the same operation as proceeds and get the same answer (158720) . 

So, the number of combinations is $2×158720=317440$. 

The total number of combinations is ${^{64}C_4}=635376$. 

Hence, the probability is,



Chinmay Katre

How is it different than the originally provided solution ?

Chinmay Katre

Sorry....I had too solved it by your method and assumed that this would be the solution provided by the site.....I better should have checked.

Edwin Gou

Dude... I just calculate the P(A)/P(B) by myself and I found out that the answer and I got the different answer. It's 620/79422 but it's not the simplest form yet but the numerator was already different.....

I'm so confused. Can anyone explain it to me why ?


I also checked the calculation and got the answer as $\dfrac{640}{1281}$. Same as the solution.

I guess you have made some calculation mistake.


lol idk anything about permutation and combinations ;) ... isnt there an easier method or something?


How about an alternate solution?

We just choose 3 random squares, 2 of which are definitely of 1 color. we just have to choose the fourth square appropriately.

If there are 3 black squares selected in the first 3, we select a white as the fourth one. if there were two white and one black in the first 3 we select a black for the fourth.

What would be the numeric solution then?