Confusion on paths and interpretation of combination Going through combinatorics as a refresher, I stumbled across a problem in my notes. 
What is the number of ways to go from the point $(0,0)$ to $(10,5)$ if the only moves you can make are to go up by one or to go right by one? 
The answer to this question is $_{15}C_5$. Of the 15 total steps you can make choose 5 of them to make as the "up" step, and the rest have to be "right" steps in order to reach $(10,5)$. 
However thinking back, isn't combination used to represent unordered sequences? In this case, the sequence "up" 5 times and then "right" 10 times is obviously different from "right" 10 times and then "up" 5 times. However, they are both the same combination sequence(unordered), so they should be counted as one path(obviously not true). 
I'm confused why combination shows up in the answer if the order in which I take my "up" steps and "down" steps matter. Is my interpretation of the problem/combination wrong?
 A: You are essentially lining up 15 items, called right_1,right_2,...,right_10,up_1,...,up_5, into a sequence to determine a route.
If you use the arrangement number, it'd be like you are assuming the two arrangements
right_1,right_2,...,right_10,up_1,...,up_5
right_2,right_1,...,right_10,up_1,...,up_5
are different, in which case you have $\ _{15}A_{15}$ different arrangements. However, they indeed result in the same route in your context, so you need to rule out the repeations. The point is right_i and right_j are identical in this problem (as well as up_i and up_j). And the final result is $$\frac{\ _{15}A_{15}}{\ _{5}A_{5}\cdot\ _{10}A_{10}}=\ _{15}C_{5}$$
Loosely speaking, combination appears in an arrangement with identical elements.
A: Choosing the positions of the up steps completely determines the sequence of moves.  
To illustrate, observe that choosing to place the up moves in positions 4, 7, 8, 12, and 15 produces the sequence $RRRURRUURRRURRU$ and the path

while choosing to place the up moves in positions 1, 6, 9, 11, and 12 produces the sequence $URRRRURRURUURRR$ and the path

Since exactly five of the fifteen moves are upwards and the rest are rightwards, there are 
$$\binom{15}{5} = \frac{15!}{5!10!}$$
possible paths from $(0, 0)$ to $(10, 5)$ in which the only permissible moves are up or to the right.  The factor of $5!$ in the denominator represents the number of ways we could permute the up moves among themselves without producing a sequence distinguishable from the given sequence; the factor of $10!$ in the denominator represents the number of ways we could permute the rightwards moves among themselves without producing a sequence distinguishable from the given sequence.
Your confusion stems from the difference between a set, in which all the elements must be different, and a multiset, in which elements are permitted to have multiplicity greater than one.   
For a set $S$ with $n$ elements, a $k$-combination is a subset with $k$ elements.  The number of such subsets is 
$$C(n, k) = \binom{n}{k} = \frac{n!}{k!(n - k)!}$$
An arrangement of a $k$-element subset of a set with $n$ elements is a $k$-permutation.  Since there are $k!$ ways to arrange the $k$ elements in the subset, the number of $k$-permutations is 
$$P(n, k) = k!C(n, k) = k! \cdot \frac{n!}{k!(n - k)!} = \frac{n!}{(n - k)!}$$
However, when we arrange a multiset, some of the elements may be the same.  A permutation of a multiset $M$ is an arrangement of the elements of the multiset in which each element appears exactly as many times as it appears in the multiset.  
In our case, the multiset is $M = \{5 \cdot U, 10 \cdot R\}$, where $U$ represents an upward move and $R$ represents a rightward move.
A particular permutation of a multiset is determined by choosing the positions of each element in the sequence.  Perhaps it would help to consider an example in which there are more than two distinct elements in the multiset.
Example.  How many anagrams does the word REPETITION have?
The ten letters of the word REPETITION form the multiset $\{1 \cdot R, 2 \cdot E, 1 \cdot P, 2 \cdot T, 2 \cdot I, 1 \cdot O, 1 \cdot N\}$.  A particular anagram of the word REPETITION is determined by choosing the positions occupied by each letter.  For instance, if we place the R in the fifth position, the E's in positions 2 and 9, the P in position 1, the T's in positions 3 and 10, the I's in positions 7 and 8, the O in position 4, and the N in position 6, we obtain the anagram PETORNIIET.
The number of such anagrams is 
$$\binom{10}{1}\binom{9}{2}\binom{7}{1}\binom{6}{2}\binom{4}{2}\binom{2}{1}\binom{1}{1} = \frac{10!}{1!2!1!2!2!1!1!}$$
since we must choose one of the ten positions for the R, two of the remaining nine positions for the E's, one of the remaining seven positions for the P, two of the remaining six positions for the T's, two of the remaining four positions for the I's, one the remaining two positions for the O, and fill the final open position with the N. 
