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If x, y, z are natural numbers, then thenumber of different solutions of theequation x +y+z-6 are:(2) 3(4) 710 |
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Answer» the natural no. are ≥ 1eg. 1,2,3,4,5.... so for x+y+z =6 the the minimum value of x,y,z can be 1 and max value would be 4. because others have to 1 so for either of x,y,z = 4 the no. of solutions are 3for either of x,y,z =3 , no. of solutions 2*3 = 6for either of x,y,z = 2 no. of solutions will be 1 , when all are 2,2,2..so total solutions are 6+3+1 = 10 |
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