Why combination with repetition doesn't use the same formula as unitary coefficient equation 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. 
 A: 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.

How many different sets of three coins if each coin can be 10 cents, 25 cents, 50 cents, or one dollar?

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 a library, there are $20$ books, each one with unlimited quantity.  How many selections of $10$ books can be done if repetitions are allowed?

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 $m$. 
