Question 290476: Respected Sir,
Can you explain with proper formula that how to count maximum possibilities of 8 different planets' situation in 12 different signs?
I know that the "formula" for the number of distinct configurations of P planets in S signs is S^P where ^ means "raised to the power of".
So, for 8 planets and 12 signs, the number of possible configurations is:
12^8, or 12*12*12*12*12*12*12*12, or 429,981,696
But
Mercury and Venus is always with the Sun. Mercury can be far maximum 1 sign before or after the Sun. Venus can be far maximum 2 Sign before or after the Sun.
Eg.
If sun is in sign 6,
Moon can be in Sign 1, Sign 2, Sign 3, Sign 4, Sign 5, Sign 6, Sign 7, Sign 8, Sign 9, Sign 10, Sign 11 or in Sign 12.
Same way, Mars, Jupiter, Saturn and Node can be far upto 12 signs from the Sun.
But in case of Mercury and Venus,
If sun is in sign 6,
Mercury and Venus cannot be in certain signs.
Mercury can be only in Sign 5, Sign 6 or in Sign 7. (Can be one sign away from Sun)
Venus can be only in Sign 4, Sign 5, Sign 6, Sign 7 or Sign 8. (Can be Two signs away from the Sun)
So we have to reduce some configurations among 429,981,696. What to do?
Can we use nCr or nPr to find out the number of configurations that we have to reduce from 429,981,696 or this can be done with the help of any other formula?
I hope you will give me the final-maximum-last Number of possible ways of planetary positions in twelve signs WITH EXPLANATION
.
Thank You
Answer by richwmiller(17219)   (Show Source):
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