N is the number of ordered pairs of non-empty sets P and Q that have the following properties:
- P ⋃ Q ={1,2,3,4,5,6,7,8,9,10,11,12}, and:
- P ⋂ Q = Φ, and:
- The number of elements of P is not an element of P, and:
- The number of elements of Q is not an element of Q
Find N.
Answer only, explanation later, if needed : 1024
Nice puzzle.
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Correction, following Daniel's remark.
I saw a very simple solution:
There are 11 couples of subsets, 1-11, 2-10,...4-8, ...11-1.
Call those couples c1,c2,...c8...c11.
For each of them 2 numbers are quantifiers (see the numbers above), 1st denoting the quantity of numbers in the 2nd set, and the 2nd quantifier denoting the quantity of numbers in the 1sd set, Those eleven are all possible ordered partitions of 12 into two addends.
The remaining 10 numbers , albeit distinct sets in distinct couples can be mapped into set (1,2,3,4,5,6,7,8,9,10)- e.g for couple c2
you have to chose 2 out of 12, i.e add 1 number chosen out of 10 to the low quantifier, the rest is thus defined to join the other quantifier.
Sum of all possible combinations is 2^10=1024 right?
wrong- there is no 6-6 partition, so comb(10,5) is the number to be subtracted: 1024-252=772 which is the correct answer.
The quick process in my mind was as follows:
2 numbers are fixed, 10 are free, therefore 1024.
The speed was ok, the accuracy not.
all in all - a very nice puzzle.
Edited on April 13, 2015, 4:12 pm
Answer only
