How many positive integer solutions are there for the equality:
$$ x_1 + x_2 + x_3 + x_4 + x_5 = 1000$$
such that: $$ x_1 < 100$$
Solution:
$$\binom{1000 - 1}{ 5 - 1} - \binom{1000 - 99 - 1}{5 - 1}$$
Now what i don't understand is:
Why the solution doesn't start as $\binom{1000}{5}$ but it starts as $\binom{1000-1}{5-1}$
Can you please give me explanation regarding this?
Thank You, Umer Selmani
Let's work with smaller numbers.
Suppose we wish to find the number of positive integer solutions to the equation $$x_1 + x_2 + x_3 + x_4 + x_5 = 10$$ A particular solution corresponds to the placement of four addition signs in the nine spaces between successive ones in a row of ten ones. $$1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1$$ For instance if we place addition signs in the second, third, fifth, and seventh spaces, we get $$1 1 + 1 + 1 1 + 1 1 + 1 1 1$$ which corresponds to the solution $x_1 = 2$, $x_2 = 1$, $x_3 = 2$, $x_4 = 2$, and $x_5 = 3$. The number of solutions is the number of ways we can select $5 - 1 = 4$ of the $10 - 1 = 9$ spaces between successive ones in a row of ten ones in which to place an addition sign, which is $$\binom{10 - 1}{5 - 1} = \binom{9}{4}$$
Since a particular solution of the equation $$x_1 + x_2 + x_3 + \cdots + x_n = k$$ in the positive integers corresponds to the placement of $n - 1$ addition signs in the $k - 1$ spaces between successive ones in a row of $k$ ones, the number of such solutions is $$\binom{k - 1}{n - 1}$$ Make sure you understand why the formula works since some authors reverse the roles of $n$ and $k$.