# Possible results that can be in a horse race

There are three horses: Uri, Uli and Buki. Results that can be possible in the race are 13.

• Uri first, Uli second, Buki third.
• Buki first, Uri second, Uli third.
• Buki first, Uri and Uli second together.
• Uri and Buki first, Uli second.
• Uri, Buki and Uli first all together.

There are eight more possible results but didn't write them all because I guess you understand what I mean. Given that information, they ask to calculate how many different possible results can be with 5 horses.

I can calculate how many results can be if all the horses come at different times:

$120 = 5\cdot4\cdot3\cdot2\cdot1$

How do I do it with the other cases? Please don't use complicated math, famous formulas or theoremas, I'm supposed to calculate this just with logic.

• For the cases of 3 horses we calculate the probabilities for the following cases and add them up: 3_0_0 : implying that all 3 horses come first, 0 come second and 0 come third Similarly, we have: 2_1_0, 1_2_0, 1_1_1 If we calculate the probabilities for each case and add them up, it sums to 13. We follow the same procedure for 5 horses. There will be of course many more cases: 5_0_0, 4_1_0, 3_2_0, 3_1_1, 2_3_0, 2_2_1, 2_1_2, 1_4_0, 1_3_1, 1_2_2, 1_1_3 Nov 21 '15 at 20:58
• @AmitSaxena Yes, I know that. I'm trying to know how's the fastest way to resolve it. Nov 21 '15 at 21:05
• There is no faster way that I can think of. I forgot to add cases where 4 and 5 places will be filled: 2_1_1_1, 1_2_1_1, 1_1_2_1, ,1_1_1_2, 1_1_1_1_1 Nov 22 '15 at 11:11
• Its rather easy e.g. the number of ways for 1_1_1_2 will be 5*4*3 = 60 Nov 22 '15 at 11:12
• Once you have solved 1_1_1_2 you don't have to solve for 2_1_1_1, 1_1_2_1 and 1_2_1_1 as they will also have 60 ways to happen. Nov 22 '15 at 11:20

If we are to use little machinery, we can divide into cases, count the number of ways for each, and then add up.

Roughly, the cases can be described as follows: $5$; $4$-$1$; $3$-$2$; $3$-$1$-$1$; $2$-$2$-$1$; $2$-$1$-$1$-$1$; $1$-$1$-$1$-$1$-$1$.

We count the number of possibilities for each.

$5$: This the the five-way tie. There is $1$ way this can happen.

$4$-$1$: Either $4$ tied first, and a loser, or $1$ winner, and $4$ tied for last, or to put it more nicely, for second. The horse by itself can be chosen in $5$ ways. It can be tied for first or tied for last, for a total of $(2)(5)$ possibilities.

$3$-$2$: Again we have $2$ possibilities, a two-way tie for first or for last. The group of $2$ can be chosen in $\binom{5}{2}=10$ ways, giving a total of $(2)(10)$.

$3$-$1$-$1$: The group of $3$ can be chosen in $\binom{5}{3}=10$ ways. It can be in any of $3$ places. The leftmost empty place can then be filled in $2$ ways, for a total of $(2)(3)(10)$.

$2$-$2$-$1$: The lone horse can be chosen in $5$ ways, and can be in any of $3$ positions. We can fill the leftmost remaining position in $\binom{4}{2}=6$ ways, for a total of $(6)(3)(5)$.

Only two to go, one of which you have done!

• What do you mean by $\bigl( \begin{smallmatrix} 5 \\ 3 \end{smallmatrix} \bigr)$ and $\bigl( \begin{smallmatrix} 4 \\ 2 \end{smallmatrix} \bigr)$? What do this symbolize? Nov 21 '15 at 21:19
• The notation $\binom{n}{k}$ is the number of ways a subset of $k$ elements can be selected from a set of $n$ elements. $$\binom{n}{k} = \frac{n!}{k!(n - k)!}$$ so $$\binom{5}{3} = \frac{5!}{3!2!} = \frac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{3 \cdot 2 \cdot 1 \cdot 2 \cdot 1} = 10$$ Nov 21 '15 at 21:20
• The binomial coefficient $\binom{n}{k}$ may be familiar to you under other symbols, such as ${}_nC_k$ or $C(n,k)$ or $C_k^n$. It is the number of ways to choose $k$ objects from a collection of $n$ objects. Nov 21 '15 at 21:25