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ankit_thawal
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Linear Algebra
Jan 13, 2018
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How to solve?
________________
Number of non-negative integer solutions such that
x+y+z=17
where x>1,y>2,z>3 is --------
system-of-equations
ankit_thawal
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Linear Algebra
Jan 13, 2018
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ankit_thawal
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Anu007
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Jan 13, 2018
edited
Jan 13, 2018
by
Anu007
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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+
3-1
C
8
=
10
C
8
= 45
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by
MiNiPanda
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Jan 13, 2018
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@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?
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Anu007
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Jan 13, 2018
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yeah...:)
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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
Navneet Kalra
answered
Jun 28, 2018
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Navneet Kalra
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