This question may rather be simple to others but I'm struggling to understand it.
So for example, in the book it says, if I want to count the number of compositions for number "$7$" (e.g $6+1,3+1+2+1$, etc..),
i) For one summand, there is only 1 composition, which is $7$ (I understand this part)
ii) For two(positive) summands, we want to count the number of integer solutions for $a_1+a_2=7$, where $a_1,a_2 > 0$ ( I understand this part because I want to count two numbers that add up to $7$ so it's $a_1+a_2=7$.)
This is equal to the number of integer solutions for $x_1+x_2=5$, where $x1,x2 \ge 0$. (I don't understand this part. How can $a_1+a_2=7$, where $a1,a2 >0$, equal to $x_1+x_2=5$, where $x1,x2 \ge 0$?? Maybe I assume the 2nd statement excludes $7$ and $0$, so it equals to $5$??)
iii) Also for three summands, we want to count the number of positive integer solutions for $y_1+y_2+y_3=7$ (I understand this part. I want to count three numbers that add up to $7$)
This is equal to the number of nonnegative integer solutions for $x_1+x_2+x_3=4$ (I don't understand how 1st and 2nd statement can be equal and why the 2nd statement add up to $4$ instead of $7$)
An easy explanation would be very much appreciated. Thank you.
Best Answer
I have $7$ identical candies. I want to distribute these among $3$ (distinct) kids so that each kid gets at least one candy.
Let $x_1$ be the number of candies Kid1 will get, let $x_2$ be the number of candies Kid2 will get, and let $x_3$ be the number of candies Kid3 will get. The number of ways to do the job is the number of solutions of $x_1+x_2+x_3=7$ in positive integers (remember, each kid gets at least $1$ candy).
Now let us think about the problem another way. Since each kid will get at least $1$ candy, give $1$ to each. There are $4$ candies left. Let $y_1$ be the number of extra candies Kid1 will get, and define $y_2$ and $y_3$ similarly. Then the $y_i$ are all $\ge 0$, but one or more may be $0$. The number of ways to distribute the $4$ extra candies is the number of solutions of $y_1+y_2+y_3=4$ in non-negative integers.
Thus the number of solutions of $x_1+x_2+x_3$ in positive integers is the same as the number of solutions of $y_1+y_2+y_3=4$ in non-negative integers.
Since formulas don't cause tooth decay, you may like formulas better than candies. Suppose that $x_1+x_2+x_3=7$, where the $x_i$ are positive integers. Let $y_i=x_i-1$. Then $y_1+y_2+y_3=(x_1-1)+(x_2-1)+(x_3-1)=7-3=4$. Moreover, all solutions of $y_1+y_2+y_3=4$ in non-negative integers are obtained from a unique solution of $x_1+x_2+x_3=7$ in positive integers by setting $y_i=x_i-1$.