3 years ago

In how many ways can the eight rooks be placed on an 8-by-8 chessboard so that no two rooks can attack one another?

Mathematics

1 answer:

saul85 [17]3 years ago

7 0

You have a total of (8!)^2/8! and that gets you. 40,320

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Bond [772]

Answer:

$R = \frac {3ST}{2S + T}$

Step-by-step explanation:

Given the equation;

$\frac {3}{R} - \frac {1}{S} = \frac {2}{T}$

Rearranging the equation, we have;

$\frac {3}{R} = \frac {2}{T} + \frac {1}{S}$

Lowest common multiple (LCM) of S and T is ST.

$\frac {3}{R} = \frac {2S + T}{ST}$

Cross-multiplying, we have;

$3ST = R(2S + T)$

Making R, the subject of formula;

$R = \frac {3ST}{2S + T}$

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