In how many ways can you bet thirteen different football matches to get exactly 12 wins?
My attempt
First I notice that there is only $1$ way to bet $13$ different matches to get $13$ wins.
Second, I notice that ther are $3^{13}$ ways to bet $13$ different football matches.
Now, I refrase the question: In how many ways can you bet thirteen different football matches to get exactly $1$ loss?
This gives
$$\binom{13}{1}\cdot2^{1} = \frac{13!}{1! (12)!}\cdot2=13\cdot2=26.$$
Now, assume you want to know in how many ways you can bet thirteen different football matches to get exactly $2$ losses?
Well that would give $$\binom{13}{2}\cdot2^{2} = \frac{13!}{2! (11)!}\cdot4=78\cdot4=312.$$
So, the general formula is $\binom{n}{r}\cdot2^{r} = \frac{n!}{r! (n-r)!}\cdot2^{r}.$
On the other hand, with $13$ matches to be bet on and exactly $1$ to be a loss, I could argue that the loosing match could be chosen in $13$ different ways, and that each match only have $2$ possible choices, as
$$13\cdot2=26.$$ This result corresponds well to the answer above.
With the same argument, with $13$ matches to be bet on and exactly $2$ to be a losses I could argue that the first loosing match could be chosen in $13$ different ways, and the second loosing match in $12$ different ways, and that each match only have $2$ possible choices, as $$13\cdot12\cdot2=312.$$
Even this result correspond to the answer above.
But, when repeating the calculation for the choice of $3$ lost games (corresponding to $10$ wins), the magic disappears:
$$\binom{13}{3}\cdot2^{3} = \frac{13!}{3! (10)!}\cdot8=286\cdot8=2288\ne13\cdot12\cdot11\cdot2=3432.$$
Where do I go wrong?
Best Answer
Your calculation of $13 \cdot 12 \cdot 11 \cdot 2$ is wrong two ways. The first three factors are the number of ways to choose the games you get wrong, but you count each order differently. If the games you lose are $ABC$ you are counting $CBA$ as different. That is the factor $3!$ between $13 \choose 3$ and $13 \cdot 12 \cdot 11 $. Then, having chosen the three games to lose, there are two ways to lose each, so the final factor should be $2^3$. The two combine to multiply $3432$ by $\frac 46$, getting $2288$.