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Quick counting problem. Find the number of solutions to x1 x2 + x3 + x4 = 12 if 0...

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nikhilnair93 | Student, Undergraduate | eNoter

Posted May 16, 2012 at 8:00 PM via web

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Quick counting problem. 

Find the number of solutions to x1 x2 + x3 + x4 = 12 if 0 =< x1 =< 2  (i.e. x1 = 0, 1 or 2)

Any help would be appreciated! 

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donatzm | College Teacher | eNotes Newbie

Posted May 21, 2012 at 7:23 AM (Answer #1)

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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.  

p(10,3)=8

p(11,3)=10

p(12,3)=12

(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.

p'(10,3)=14

p'(11,3)=16

p'(12,3)=19

 (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

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