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How many solutions are there for the equation $x+y+z+u=29$ subject to the constraints that $x \geq 1, \: \: y \geq 2 \: \: z \geq 3 \: \: and \: \: u \geq 0$?

4960

2600

23751

8855
recategorized | 865 views

x+y+z+u = 29 this is combination with repetition...

x>=1,y>=2,z>=3,u>=0 so allocate this to directly..

29-(1+2+3+0) = 23

(23+4-1)C23 = 2600

This problem is similar to distributing 29 similar ball into 4 numbered boxes..
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29 - 6= 23

(23 + 3) C3 = 26C3 = 2600