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Find the number of ways 5 dices can be rolled to get a sum of 25.

While solving this question, the way we solve it is $x_1+x_2+x_3+x_4+x_5$ $=25$ where $1<=x_i<=6$

So we replace $x_i$ by $y_i =6-x_i$ , which is $x_i=6-y_i$

substituting $x_i$ in the above equation we get it as→ $(6*5) – (y_1+y_2+y_3+y_4+y_5)$ $=25$

$(y_1+y_2+y_3+y_4+y_5)$ = $5$

After solving this equation by integer solutions formula $(n-r+1)! / (n! * (r-1)!)$ we get the ans as → $126$


Now consider this problem,

The number of non-negative integer solutions such that $x_1+x_2+x_3=17$ where $x_1>1, x_2>2 , x_3>3$ is ___________________

While solving this we are solving it like → $y_1= x_1-2$ , $y_2=x_2 -3$ , $y_3=x_3-4$

so, $x_1= y_1+2$ , $x_2=y_2 +3$ , $x_3=y_3+4$

Now we substitute this in our original equation to get→

$ y_1+2+y_2 +3+y_3+4 =17$

$ y_1+y_2 +y_3 =8$

and after solving this we get the ans as $45$


Now I have a $DOUBT$ here, in the second problem since when $x_1>1$ we make it as $x_1 = y_1+2$ , but in the first problem all the dices should have value $>0$ , so why in that case we haven’t made $x_i=y_i+1$ for all the cases?


And moreover what if the question was like

$x_1+x_2+x_3=12$   ,  $2<=x<=5$ then how to solve this using integer solution and applying the formula $(n-r+1)! / (n! * (r-1)!)$ ?

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  • $\begingroup$ In the first case, it's a way to incorporate the constraint that each die is at most 6. The right hand side you end up with is 5 which is less than 6 $\endgroup$ – ogogmad Oct 19 '19 at 18:28
  • $\begingroup$ In your third case, I would write $y_i = 3 + x_i$, use the formula, and then take away 3 solutions $\endgroup$ – ogogmad Oct 19 '19 at 18:29
  • $\begingroup$ Alternatively, in the third case you could write $x_i = 5 - y_i$, simplify to $y_1 + y_2 + y_3 = 3$, and then observe that you don't have to take away solutions $\endgroup$ – ogogmad Oct 19 '19 at 18:31
  • $\begingroup$ In my first example, I meant to use $2$ not $3$. $\endgroup$ – ogogmad Oct 19 '19 at 18:33
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We have a lot of freedom in how to solve the third problem.

We could substitute $x_i = 2 + y_i$, use the formula, and then take away three solutions. Those three solutions correspond to $(6,0,0), (0,6,0)$ and $(0,0,6)$, which violates the constraint that $x \leq 5$.

Alternativrly, we could substitute $x_i = 5 - y_i$, simplify the equation to $y_1 + y_2 + y_3 = 3$, and then use the counting formula. Observe that we don't need to take away solutions, because the constraint that $2<=x<=5$ is automatically satisfied. This is slightly more convenient.

So the second approach is marginally better in that it doesn't require taking away solutions. But the advantage either way is negligible.

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  • $\begingroup$ Thanks for your explanation, couldn't upvote because of low points.. $\endgroup$ – Turing101 Oct 19 '19 at 19:13
  • $\begingroup$ Moreover I have a doubt here, say if the equation was $x_1+x_2+x_3=12$ , $1<=x<=5$, then what should have been our approach as $x_i=5-y_i$ would give the same result here also, but which is not the case, here we should have more number of solutions $\endgroup$ – Turing101 Oct 19 '19 at 19:14
  • $\begingroup$ None of $x_1$, $x_2$ or $x_3$ can be $1$, as then the sum is always less than 12. So the number of solutions does indeed stay the same $\endgroup$ – ogogmad Oct 19 '19 at 19:27
  • $\begingroup$ okay, and if it was like $x1+x2+x3=12 , 3<=x<=5$, then what? aren't there any generalized formulas for these problems? $\endgroup$ – Turing101 Oct 20 '19 at 3:56
  • $\begingroup$ If $3 \leq x \leq 5$ then you have to take away solutions. There's no general formula $\endgroup$ – ogogmad Oct 20 '19 at 4:04

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