Test Series ACE

Question

How to solve?
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Number of non-negative integer solutions such that

x+y+z=17

where x>1,y>2,z>3 is --------

Comments

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+3-1C₈ = ¹⁰C₈ = 45

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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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yeah...:)

Answer 1

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