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Question: Three squares are chosen on a chess board, the chance that they should be in a diagonal line is \...

Three squares are chosen on a chess board, the chance that they should be in a diagonal line is
A)7744A)\dfrac{7}{{744}}
B)5744B)\dfrac{5}{{744}}
C)10744C)\dfrac{{10}}{{744}}
D)11744D)\dfrac{{11}}{{744}}

Explanation

Solution

We will make use of the probability formula to solve this problem, and in a chessboard, there are a total 6464 squares and out of that sixty-four squares we have to choose only three squares.
In that 6464 square, there are 1616 diagonal lines on the chessboard.
We need to choose three squares and find the probability that three squares lie in that diagonal line.

Formula used:

P=FTP = \dfrac{F}{T} where P is the overall probability, F is the possible favorable events and T is the total outcomes from the given.
ncr=n!r!(nr)!{}^n{c_r} = \dfrac{{n!}}{{r!(n - r)!}} which is the combination formula to find the number of ways.

Complete step-by-step solution:
Given that we have Three squares chosen on a chessboard, we need to find the chance that they should be in a diagonal line.
Let the total number of outcomes will be equal to the number of ways of choosing 33 squares out of 6464 square which is present in the chessboard is 64C3{}^{64}{C_3} (by using the method of combination which is to find the number of ways)
Also, the possible number of outcomes will be equal to the number of ways of choosing 33 square out of 1616 diagonal lines in the chessboard is 16C3{}^{16}{C_3}
Thus, the favorable event is 16C3{}^{16}{C_3} and the total event is 64C3{}^{64}{C_3}
Hence, we get P=FT=16C364C3P = \dfrac{F}{T} = \dfrac{{{}^{16}{C_3}}}{{{}^{64}{C_3}}}.
Since the combination formula is ncr=n!r!(nr)!{}^n{c_r} = \dfrac{{n!}}{{r!(n - r)!}}
Further solving we have, P=16C364C3=16!3!13!64!3!61!P = \dfrac{{{}^{16}{C_3}}}{{{}^{64}{C_3}}} = \dfrac{{\dfrac{{16!}}{{3!13!}}}}{{\dfrac{{64!}}{{3!61!}}}}
Taking the inverse division, P=16!3!13!64!3!61!=16!3!13!×3!61!64!P = \dfrac{{\dfrac{{16!}}{{3!13!}}}}{{\dfrac{{64!}}{{3!61!}}}} = \dfrac{{16!}}{{3!13!}} \times \dfrac{{3!61!}}{{64!}}
Canceling the common terms, P=16!13!×61!64!=16.15.14.13!13!61!64.63.62.61!16.15.1464.63.62P = \dfrac{{16!}}{{13!}} \times \dfrac{{61!}}{{64!}} = \dfrac{{16.15.14.13!}}{{13!}}\dfrac{{61!}}{{64.63.62.61!}} \Rightarrow \dfrac{{16.15.14}}{{64.63.62}}
Hence, we get P=16.15.1464.63.6210744P = \dfrac{{16.15.14}}{{64.63.62}} \Rightarrow \dfrac{{10}}{{744}} a number of ways that the three squares are diagonal lines
Therefore, the optionC)10744C)\dfrac{{10}}{{744}} is correct.

Note:
1. We used the basic concept of factorial in the combination method to solve further, which is n!=n(n1)(n2)...2.1n! = n(n - 1)(n - 2)...2.1
2. Probability is the term mathematically with events that occur, which is the number of favorable events that divides the total number of the outcomes.
3. If we divide the probability and then multiplied with the hundred then we will determine its percentage value.
4. 16\dfrac{1}{6} which means the favorable event is 11 and the total outcome is 66