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Question

Find number of negative integral solution of equation x+y+z=12

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Solution

x+y+z=12 where x,y,z are negative integers.
12=1+1+10;1+2+9;1+3+8;1+4+7;1+5+6;here if x=1,y+z can have 5 different values but for example y=1,z=10.Also z=1,y=10 is also possible.
So, with x=1 we can have 10 possibilities
Similarly if y=1,10 possibilities and if z=1,10 possibilities.
If we take x=2 or any even number we will get y=5 and z=5 so we can't reverse so for x=even we can have only 9 possibilities.
Now,y=even can also have only 9 possibilities and z=even can also have only 9 possibilities.So, for even there are 27 possibilities.
xodd(1,3,5,7,9,11)
we get 6 odd cases
6×30=180possibilities ( each odd case has 30 possibilities.
When xeven(2,4,6,8,10) we hve 5 even cases.
5×27=135 possibolities.
Now if x=12,y=z=0 or if y=12 or z=12 there are no possibiliites because 0 is not a negative integer.
Total positive integers set for x+y+z=12 is 180+135=315
So, for x+y+z=12 there will be 315 negative integral solutions.

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