Distributing Identical Objects into Distinct Cells Inequality vs Strict Equality So, I understand the formula for any general instance is $C(r - 1,n - 1)$ where $r$ is the number of spaces and $n$ is the number of distributions, but how does strict equality versus inequality affect this formula?
For instance, is my reasoning for the two problems below correct?
$x_1 + x_2 + x_3 + x_4 + x_5 = 63$, all $x_i \geq 0$
Use the inflation principle such that ($62$ spaces $+ 5 = 67$, $4$ partitions)
Therefore, the answer is $C(67,4)$
$x_1 + x_2 + x_3 + x_4 + x_5 \leq 63$, all $x_i \geq 0$
Use the inflation principle such that ($62$ spaces $+ 5 = 67$, $4$ partitions)
Therefore, the answer is $C(67,4)$ 
This however, does not make sense to me. Shouldn't the inequality have more combinations?
 A: Your suspicion is correct.
You are correct that the number of non-negative integer solutions to the equation
$$x_1 + x_2 + x_3 + x_4 + x_5 = 63$$
is $\binom{67}{4}$.
We wish to find the number of solutions of the inequality
$$x_1 + x_2 + x_3 + x_4 + x_5 \leq 63 \tag{1}$$
in the non-negative integers.  Let
$$s = 63 - (x_1 + x_2 + x_3 + x_4 + x_5)$$
Then $s$ is a non-negative integer.  Moreover,
$$x_1 + x_2 + x_3 + x_4 + x_5 + s = 63 \tag{2}$$
is an equation in the non-negative integers that is equivalent to inequality 1.  A particular solution of equation 2 corresponds to the placement of five addition signs in a row of $63$ ones.  For instance,
$$1 1 1 1 1 1 1 1 1 1 + 1 1 1 1 1 1 1 1 1 1 1 1 + 1 1 + 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 + 1 1 1 1 1 1 1 + 1 1 1 1 1 1 1 1 1 1 1 1$$
corresponds to the solution $x_1 = 10$, $x_2 = 12$, $x_3 = 2$, $x_4 = 20$, $x_5 = 7$, and $s = 12$ of equation 2 and $x_1 = 10$, $x_2 = 12$, $x_3 = 2$, $x_4 = 20$, and $x_5 = 7$ of inequality 1.  Thus, the number of non-negative integer solutions of equation 2 and, equivalently, inequality 1 is
$$\binom{63 + 5}{5} = \binom{68}{5}$$
since we must choose which five of the $68$ symbols (five addition signs and $63$ ones) will be addition signs.
