There are 15 people and 4 positions (north, south, east, and west)
How can I evaluate how many possible arrangements of people can be done where position matters? Having only one person per a position at a time.
[Math] In how many ways 15 people be chosen for 4 distinct positions
combinatorics
Related Solutions
Since $n$ must be even in order for the walk to be closed, I’ll write $n=2m$. Let $n_N,n_S,n_E$, and $n_W$ be the number of steps north, south, east, and west, respectively. Then $n_N=n_S$ and $n_E=n_W$, so $n_N+n_E=n_S+n_W=m$ and $n_N+n_W=n_S+n_E=m$.
Note also that if we choose $n_N,n_S,n_E$, and $n_W$ so that $n_N+n_E=n_S+n_W=m$ and $n_N+n_W=n_S+n_E=m$, then automatically $n_N=n_S$ and $n_E=n_W$.
Now imagine charting an $n$-step walk by starting with a strip of $n$ squares and labelling each square N, S, E, or W for steps north, south, easy, and west, respectively. However, we’ll do it in a slightly odd way. First choose any $m$ squares and mark them $\nearrow$. These will be the $m$ steps that go either north or east. Then choose any $m$ squares and mark them $\nwarrow$; these will be the $m$ steps that go either north or west. Clearly this procedure can be carried out in $\binom{2m}m^2$ ways. Now go through the strip and change the markings according the following rules:
- a square marked with both $\nearrow$ and $\nwarrow$ is marked N.
- a square marked only with $\nearrow$ is marked E.
- a square marked only with $\nwarrow$ is marked W.
- an unmarked square is marked S.
It follows from the remark in the second paragraph that we’ve laid out a chart for a path that returns to the origin, and it’s not hard to see that every $n$-step path that returns to the origin has a chart that can be produced in this way. There are $\binom{2m}m^2$ charts produced in this way, so there are $\binom{2m}m^2$ $n$-step paths that return to the origin.
Well, consider the positions available. There are three. They literally told you that they are different. So consider positions one two and three.
You have five candidates. You fill the first position and there are 5 ways to do it.
Then you consider the second position. You already hired one person, so there are 4 people left. Hence, there are four ways to fill position two.
Lastly, there are 3 ways to fill spot three.
Hence, there are $5\cdot 4\cdot 3$ ways to fill the positions.
Secretly, we used permutations. There are are 3 positions that are different and hence "the order matters". There are five candidates. Hence ${_5\mathsf P}_3$.
Best Answer
$\binom{15}{4}$ - it's called a binomial coefficient, and it's the number of combinations C(15,4).