Three 1x1 squares are taken on a 8x8 chess board.

1. What is the probablity, that all three squares are diagonal to each other ?

2.What is the probability, that all three squares are diagonal to each other and lie adjacent.

what are the answers????

answer1:

3612/64c3

72/64c3

Can yOU eXPLAIN hOW?

think the answer of first Q is = 392/64_C_3

Explaination-->

any 3 square in the same diagonal if the squares selected from the diagonal of the chess board....

1st and 2nd diagonal rejected becoz in this there are less than 3 box.

there are Total 15 diagonal so due to 1st and 2nd 4 diagonals(consider both side) has been rejected ..

so remaining diagonals are 3c3 , 4c3 , 5c3 , 6c3 , 7c3 , 8c3 --all are repeated except 8c3.

all r again repeted from other SIDE ......(consider from UPPER RIGHT corner and UPPER LEFT corner)

so finally ---> 2*( 2*(3c3 + 4c3 + 5c3 + 6c3 + 7c3) + 8c3) == 392

similarly.......

==> 2*( 2*( 1 + 2 + 3 + 4 +5) + 6 ) == 72

so answer is 72/64c3

First consider left to right

One large diagonal: 8c3

The rest are repeated above n below the diagonal i.e 7c3 + 6c3 + 5c3 + 4c3 + 3c3

Therefore,

Total number of squares that lie on (left to right) diagonals is

(7c3 + 6c3 + 5c3 + 4c3 + 3c3)*2 + 8c3

But we need to consider the right to left diagonals as well... and due to symmetry we can simple multiply the above term with 2

Thus total such squares (sharing a diagonal)= 2*((7c3 + 6c3 + 5c3 + 4c3 + 3c3)*2 + 8c3)

Thus the probability is 392/64c3

anymore answers??

## Sunday, April 26, 2009

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