Unitary coefficient equation uses the formula
$$ x_1 + x_2 + \ldots + x_k = m $$
$$ C(m + k – 1, k – 1) $$
(considering $x$ can be $0$, i.e solutions in positive integers)
But for combination with repetition, my book says that the formula is $C(m + k – 1, k)$ (note it is not $k-1$) and says that it is equivalent to proposing the problem as:
$$x_1 + x_2 + \ldots + x_k = m$$
The weird thing is that in the example for combination with repetition is exactly the way of a equation with unitary coefficients.
Example: How many different sets of three coins can be made if each coin can be 10, 25, 50 cents or 1 dollar.
It says that this can be thought as $x_1 + x_2 + x_3 + x_4 = 3$
In this case is $C(6, 3)$ which seems to correlate with the first theorem.
But another example says "In a library there 20 math books each one with unlimited quantity. How many elections of 10 books can be done if repetitions are allowed?"
And puts as answer $C(29, 10)$ instead of $C(29, 9)$
As my thinking is that it should be $C(20 + 10 – 1, 10 – 1)$
I am a bit confused.
Idiot-proof answer is much appreciated.
Best Answer
The number of solutions of the equation $$x_1 + x_2 + x_3 + \ldots + x_k = m$$ in the nonnegative (not positive) integers is $$\binom{m + k - 1}{k - 1} = \binom{m + k - 1}{m}$$ The author of your book seems to be using the formula $$\binom{m + k - 1}{m}$$ in which case $\binom{m + k - 1}{k}$ is a rather unfortunate typographical error.
There are four different types of coins, so $k = 4$. A total of three coins are selected, so $m = 3$. Thus, we obtain the equation $$x_1 + x_2 + x_3 + x_4 = 3$$ Our formula yields $$\binom{m + k - 1}{k - 1} = \binom{3 + 4 - 1}{4 - 1} = \binom{6}{3} = \binom{3 + 4 - 1}{3} = \binom{m + k - 1}{m}$$
In this case, there are $20$ types of books, so $k = 20$. A total of $10$ books are selected, so $m = 10$. Our formula yields $$\binom{m + k - 1}{k - 1} = \binom{10 + 20 - 1}{20 - 1} = \binom{29}{19} = \binom{29}{20} = \binom{10 + 20 - 1}{20} = \binom{m + k - 1}{m}$$ where $$\binom{29}{19} = \binom{29}{10}$$ since $$\binom{n}{j} = \binom{n}{n - j}$$
Your confusion stems from what appears to be a typographical error in the text and, in the second problem, from distinguishing between $k$ and $m$.