- Apr 28, 2020
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There is simpler solution.That's an interesting math problem actually.
Q: We have 5 girls in an orgy, every girl can be or not to be pregnant. How many renders Tinkerer needs per image to cover all possible combination?
A:
No-one is pregnant. Obviously 1.
1 girl is pregnant. That's an easy one as well - 5.
2 girls are pregnant. A bit more complicated. 5!/(5-2)! 2! = 5!/3! 2! = 10
3 girls are pregnant. Basically the same as previous. 5!/(5-3)! 3! = 5!/2! 3! = 10
4 girls are pregnant. 5!/(5-4)! 4! = 5!/1! 4! = 5
All girls are pregnant. 1 again ofc.
Final answer: Tinkerer need 1+5+10+10+5+1 = 32 renders per image
If you add 3rd factor (a-bit-pregnant for example) it's getting wayyyy more complicated ofc. But if you add Maghda the answer will always be 0. At least for me.
Somehow I doubt they will ever ask that question in math class.
Number of renders is 2^n where n is a number of girls, so for 5 girls it's 2^5=32. In case of 3 factors we would have 3^n, so for 5 girls it's 3^5=243. Your solution (sum of six Newton's symbols) is also correct but too complicated.
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