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3 years ago

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Eights rooks are placed randomly on a chess board. What is the probability that none of the rooks can capture any of the other r

ooks? Translation for those who are not familiar with chess: pick 8 unit squares at random from an 8 × 8 square grid. What is the probability that no two chosen squares share a row or a column? Hint. You can think of placing the rooks both with or without order, both ap- proaches work.

1 answer:

lbvjy [14]3 years ago

4 0

Answer:

$4.51\times10^{-10}$

Step-by-step explanation:

There are only 2 solutions for none of the rooks can capture any of the other rooks: only if they are laid out in either of the 2 diagonal of the chess board

The number of possible combination for the rock to laid out in 8x8 (64 slots) chess board

$\frac{64 * 63 * 62 * 61 * 60 * 59 * 58 * 57}{8!} = \frac{1.78\times10^{14}}{40320} = 4426165368$ possible combination

So the probability is pretty thin

$P = 2 / 4426165368 = 4.51\times10^{-10}$

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<em>The inverse of $y=3^x$ is </em>

$\boxed{\boxed{y=\log_3 x}}$

<u>Solution-</u>

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$y=3^x$

We can get the inverse by interchanging he variable x and y among themselves and then separating each variables.

So in the inverse would be,

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