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$2$ squares $(1\times 1)$ are chosen at random from a chess board,

Then find the number of ways in which $2$ chosen square have neither any side nor any corner in common.

$\bf{Attempt}$ Number of ways of choosing squares is $\displaystyle \binom{64}{2}$

Number of ways in which exactly one side is common is $\displaystyle 2(7\times 7) = 98$

Now how can I calculate for a corner in common?

Thanks.

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    $\begingroup$ Two squares are chosen at random out of what? From the use of $64$ maybe you mean only $1 \times 1$ squares out of an $8 \times 8$ grid. Wouldn't it be better to explain that? $\endgroup$ – Ross Millikan Jun 20 '17 at 2:16
  • $\begingroup$ yes Ross Millikan $\endgroup$ – DXT Jun 20 '17 at 2:18
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The cute way to compute the number of pairs that touch at only a corner is that the corners they can touch at must be one of the internal points. For each internal point there are two pairs that can touch there, so there are .....?

Your computation of the number of edge meetings is not correct.

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Hint: There are three possible ways to pick the first square; corners, edges, and center. Corners remove 4 squares from the field, edges remove 6 squares from the field and centers remove 9 squares from the field. Now all you need to know is the probability of each and add them together.

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