2
$\begingroup$

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 $(n-r+1)! / (n! * (r-1)!)$ ?

$\endgroup$
4
  • $\begingroup$ 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 $\endgroup$ Commented Oct 19, 2019 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 $\endgroup$ Commented Oct 19, 2019 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 $\endgroup$ Commented Oct 19, 2019 at 18:31
  • $\begingroup$ In my first example, I meant to use $2$ not $3$. $\endgroup$ Commented Oct 19, 2019 at 18:33

1 Answer 1

Reset to default
0
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ Thanks for your explanation, couldn't upvote because of low points.. $\endgroup$ Commented Oct 19, 2019 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 $\endgroup$ Commented Oct 19, 2019 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 $\endgroup$ Commented Oct 19, 2019 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? $\endgroup$ Commented Oct 20, 2019 at 3:56
  • $\begingroup$ If $3 \leq x \leq 5$ then you have to take away solutions. There's no general formula $\endgroup$ Commented Oct 20, 2019 at 4:04

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.