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Quick counting problem. Find the number of solutions to x1 x2 + x3 + x4 = 12 if 0...
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Are you missing a plus sign between x1 and x2?
If so, we can try solving it this way:
Think of the problem in three parts. How many ways can we write 12 as the sum of three integers (x1=0), how many ways can we write 11 as the sum of three integers (x1=1), and how many ways can we write 10 as the sum of three integers (x1=2)?
Can x2, x3, and x4 be any integer? Only positive? Anything non-negative?
Assuming that x2, x3, and x4 must be positive numbers, then we need to find the number of partitions of 10, 11, and 12 into exactly 3 parts.
(Where p(n,k) is the number of ways to write n as the sum of exactly k numbers)
So there are 8+10+12 = 30 solutions to x1+x2+x3+x4=12 with x2,x3,x4 all positive and 0<=x1<=2.
Assuming x2,x3, and x4, can be 0 or positive, then we have more solutions. We need to find the number of partitions of 10,11, and 12 into at most 3 parts.
(Where p'(n,k) is the number of ways to write n as the sum of at most k numbers)
So there are 14+16+19=49 solutions to x1+x2+x3+x4=12 with x2,x3,x4 all 0 or positive, and 0<=x1<=2
Posted by donatzm on May 21, 2012 at 7:23 AM (Answer #1)
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