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  • If I have $a_1+a_2 \leq 10$, with $a_1, a_2 \in \{0, \, 1, \, 2, \, \cdots, \, 10 \}$:

To count the number of possible combinations for $a_1$ and $a_2$ such that $$a_1+a_2 \leq 10\quad\mbox{and}\quad a_1, a_2 \in \{0, \, 1, \, 2, \, \cdots, \,10 \} $$ we need the formula $\sum_{a_1=0}^{10}\sum_{a_2=0}^{10-a_1}\, 1$.

Note that I need that the result of each combination to be of the form $e^{a_1 u + a_2 d}$ with a fixed $u>0$ and $d<0$.

  • If I have $\sum_{i=1}^{n} a_i \leq 10$, with $n>1$ and $a_1, a_2, \, \cdots, a_n \in \{0, \, 1, \, 2, \, \cdots, \, 10 \}$.

Is there a mathematical method or a reference to modify the aforementioned formula?

Any help will be very appreciated !.

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  • $\begingroup$ Can you see why the inner sum is just $10-a_i$? $\endgroup$ Commented May 15 at 12:32
  • $\begingroup$ Consider stars and bars method: en.wikipedia.org/wiki/Stars_and_bars_(combinatorics) $\endgroup$ Commented May 15 at 12:34
  • $\begingroup$ @preferred_anon Thank you for your comment. Because I need all the result combinations to be of the form $e^{a_1 u + a_2 d}$ with a fixed $u>0$ and $d<0$. $\endgroup$ Commented May 15 at 12:41

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I'm a JEE aspirant and these types of questions are quite common to us. The method I use in permutation and combination questions is trying to imagine what I am doing.

To explain this in a simpler way let us look at an example problem: $x+y+z \geqslant 150$ such that $0 \leqslant x,y,z \leqslant 60$.

To solve this question we could firstly notice that the sum $x+y+z$ can be $180$ at maximum. So let us simply assume that $x+y+z-p=150$ ($p$ will take values such that this sum is always $150$). Now we could write $x=60-x_1$, $y=60-y_1$ ,$z=60-z_1$ (This necessarily doesnt change how $x$,$y$ and $z$ are choosen, it just helps all the varibales get the same sign). Substituting in our original equation we get $30=x_1+y_1+z_1+p$. Now this is the same as distributing $30$ identical chocolates among $4$ kids( with no restriction on how many chocolates an individual child recieves). This is simply just 33C3. But if we had been given a restriction that $x$ and $y$ must be $\geqslant 1$ and $z \geqslant 0$. Then we would simply use a previous equation $30=x_1+y_1+z_1+p$ and forcefully give $1$ chocolate to $x$ and $1$ chocolate to y (which means $x=59-x_1$ as we already gave it a chocolate). So our new equation would be $28=x_1+y_1+z_1+p$. This would just equate to 31C3.

In your example question of $a_1+a_2 \leqslant 10$, we could simply introduce a new variable p again such that $a_1+a_2+p=10$. Now this is the same as distributing $10$ identical chocolates to $3$ children with no resitions on how many chocolates an individual child recieves. Similarly for the case where you asked $a_1+a_2+ \dots +a_n \leqslant 10$, you can just introduce a new variable $p$ such that $a_1+a_2+\dots+a_n+p=10$ and similarly distribute chocolates among children. Hope this helps you.

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  • $\begingroup$ Hi, here is the tutorial of how to format math equations on this website: math.meta.stackexchange.com/questions/5020/… $\endgroup$ Commented May 15 at 13:00
  • $\begingroup$ @RupertRybka thanks so much I was new and really didn't know how to represent symbols. $\endgroup$ Commented May 15 at 13:03
  • $\begingroup$ Great, thank you a lot for your answer. $\endgroup$ Commented May 16 at 8:29
  • $\begingroup$ @LisztMorero Just happy that I could help $\endgroup$ Commented May 16 at 12:15

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