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Number of non-negative integer solutions such that

x+y+z=17

where x>1,y>2,z>3 is --------
in Linear Algebra by Active (1.4k points) | 117 views
0
Is it 45.
0
how to solve?
plzz elaborate.
+2

x+y+z=17
where x>1,y>2,z>3 

x+2+y+3+z+4= 17

x+y+z= 17- 9

x+y+z= 8

8+3-1C= 10C8 = 45

0
@Anu007 there is greater than sign not greater than equal to, so x must be at least 2 , y at least 3 and z at least 4 right?
0
yeah...:)

1 Answer

0 votes
since x>1,y>2 and z>3

so new equation will be

x+1+y+3+z+4=17

=>x+y+z=8

now no. of solution of these equations are (n+r-1)C(r-1)

here n=8 r=3(no. of unknowns)

so 10C2=45 Ans
by Active (1.4k points)

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