in Combinatory recategorized by
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7 votes
7 votes
Count the number of non-negative integer solutions to
$$
3 x_{1}+3 x_{2}+3 x_{3}+7 x_{4}=22 .
$$
in Combinatory recategorized by
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1 Answer

8 votes
8 votes

We divide into cases. The only possible values for $x_{4}$ are $x_{4}= 0,1,2,3$, since otherwise the left hand side is too large.

  • Case $x_{4}=0:$ We get $3\left(x_{1}+x_{2}+x_{3}\right)=22$. No solutions as the right hand side is not a multiple of $3$
  • Case $x_{4}=1:$ We get $3\left(x_{1}+x_{2}+x_{3}\right)=15$, so $x_{1}+x_{2}+x_{3}=5$ which has $\left(\begin{array}{l}7 \\ 2\end{array}\right)=21$ solutions.
  • Case $x_{4}=2:$ We get $3\left(x_{1}+x_{2}+x_{3}\right)=8$, no solutions.
  • Case $x_{4}=3:$ We get $3\left(x_{1}+x_{2}+x_{3}\right)=1$, no solutions.

Therefore, there are $21$ solutions in total.

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edited by

4 Comments

@minip 

$x_4=4$ is possible if you break the terms like that but it will not give the non-negative integer solutions. I am mentioning the reason below.

As you have broken terms, you could write $3(x_1 + x_2 + x_3 + x_4) + 4x_4 =22 \implies x_1+x_2+x_3+x_4 = \frac{22-4x_4}{3}$

Since, left side is non-negative integer, right side should also be integer and non-negative which is possible when $x_4=1,4$

When $x_4 = 1$ then our equation becomes $x_1+x_2+x_3+1 = \frac{22 – 4\times 1}{3} \implies x_1 + x_2+x_3 =5$  

When $x_4 = 4$ then our equation becomes $x_1+x_2+x_3+4 = \frac{22 – 4\times 4}{3} \implies x_1 + x_2+x_3 =-2$

$ x_1 + x_2+x_3 =-2$ can not produce non-negative integers $x_1,x_2,x_3$ such that their sum will be a negative number.

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@shishir__roy 

@ankitgupta.1729 

Thanks a lot for your patience. I made a blunder, I understood.

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@shishir__roy

@ankitgupta.1729

Anybody would have GIVEN UP/moved over me, seeing my blunder & upon that, my cross-arguments. That’s why I am again thanking for your patience & brotherhood.

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