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| Dec 18, 2006 7:55 am |
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re: re: Riddle |
H.K.L. Sachdeva
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Going through the old ones, I find that the maximum possible
number of ways the nine planets may not occupy the 6th,
8th & 12th house, has to be calculated as follows :-
1. Seven Planets (other than Rahu & Ketu) can occupy any
place but Rahu and Ketu are always in the opposite house
to each other.
2. For the seven Planets (other than Rahu & Ketu) not to
occupy the 6th, 8th or 12th house means they can occupy
any of the other nine houses, the probability of which is
9 for each planet - meaning - 63 (9 x 7).
3. Now for Rahu & Ketu, their positions if they occupy
6th, 8th or 12th houses, will be like (For Rahu) 6 - 12,
8 - 2 & 12 - 6 or the other way round (For Ketu) 12 - 6,
2 - 8 or 6 - 12 and we see that 6 - 12 & 12 - 6 combinations
are repeated twice. So out of total of these six positions,
we remove two and we are left with a total of four positions.
Thus to calculate the probability of Rahu occupying the
other eight houses (Ketu will be automatically taking the
opposite house) will be 8 (8 x 1).
4. So the total number of possible combinations for none
of the planets to be there in 6th, 8th & 12th houses will
be 63 + 8 = 71 [out of a total of 96 (12 x 8) - taking
only one set of combinations for Rahu & Ketu].
Private Reply to H.K.L. Sachdeva (new win) |
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