Question

Four squares are chosen at random on a chess board. If the probability that they lie on a diagonal line is $\frac{\lambda}{{}^{64}C_4}$, then the value of $\lambda$ must be

Answer 1

364

Answer 2

To find the value of $\lambda$, we need to calculate the total number of ways to choose four squares on a chessboard such that they lie on a diagonal line. A standard chessboard is an 8x8 grid.

1. Total number of ways to choose 4 squares from 64: The total number of ways to choose 4 squares from 64 squares on a chessboard is given by the combination formula: $N_{total} = {}^{64}C_4$

2. Number of ways to choose 4 squares such that they lie on a diagonal line: For four squares to lie on a diagonal line, the diagonal line must be at least 4 squares long. We need to count the number of such diagonal lines on an 8x8 chessboard and then for each line, calculate the number of ways to choose 4 squares from it.

There are two types of diagonals: those running from top-left to bottom-right (or vice-versa, i.e., constant difference of coordinates) and those running from top-right to bottom-left (or vice-versa, i.e., constant sum of coordinates). For an $n \times n$ board (here $n=8$), the number of diagonals of a certain length are:

• Length 8: There are 2 such diagonals (the main diagonals, e.g., A1-H8 and A8-H1).
• Length 7: There are 4 such diagonals (e.g., A2-G8, B1-H7, etc.).
• Length 6: There are 4 such diagonals.
• Length 5: There are 4 such diagonals.
• Length 4: There are 4 such diagonals.

Now, we calculate the number of ways to choose 4 squares for each length:

• Case I: Diagonals of length 8 There are 2 diagonals of length 8. From each, we choose 4 squares. Number of ways = $2 \times {}^{8}C_4 = 2 \times \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 2 \times 70 = 140$.

• Case II: Diagonals of length 7 There are 4 diagonals of length 7. From each, we choose 4 squares. Number of ways = $4 \times {}^{7}C_4 = 4 \times \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 4 \times 35 = 140$.

• Case III: Diagonals of length 6 There are 4 diagonals of length 6. From each, we choose 4 squares. Number of ways = $4 \times {}^{6}C_4 = 4 \times \frac{6 \times 5}{2 \times 1} = 4 \times 15 = 60$.

• Case IV: Diagonals of length 5 There are 4 diagonals of length 5. From each, we choose 4 squares. Number of ways = $4 \times {}^{5}C_4 = 4 \times 5 = 20$.

• Case V: Diagonals of length 4 There are 4 diagonals of length 4. From each, we choose 4 squares. Number of ways = $4 \times {}^{4}C_4 = 4 \times 1 = 4$.

3. Total number of favorable outcomes: Summing the ways from all cases: $N_{favorable} = 140 + 140 + 60 + 20 + 4 = 364$.

4. Calculate the probability: The probability that the four chosen squares lie on a diagonal line is: $P = \frac{N_{favorable}}{N_{total}} = \frac{364}{{}^{64}C_4}$

5. Determine the value of $\lambda$: Given that the probability is $\frac{\lambda}{{}^{64}C_4}$, we can compare the two expressions: $\frac{\lambda}{{}^{64}C_4} = \frac{364}{{}^{64}C_4}$ Therefore, $\lambda = 364$.