In the following figure, on the left, it is represented in an o.n. Oxyz, a regular octahedron [ABCDEF], whose vertices belong to the coordinated axes. Assume that the [ABC] face of the octahedron is numbered with the number 1, as shown in the figure on the right. It is intended to number the octahedron's remaining faces with numbers from 2 to 8 (a different number in each face).
How many different ways can the remaining seven faces be numbered, so that, after the octahedron have all the faces numbered, at least three of the concurrent faces at vertex A are numbered with odd numbers?
I've tried to solve this problem with the following reasoning:
$${}^{3}\textrm{C}_{2}*(^{3}\textrm{P}_{2})*5! = 2160$$
Where:
${}^{3}\textrm{C}_{2}$ is the number of ways we can select the 2 faces that will have odd numbers from the 3 faces that are concurrent at vertex A;
$_{}^{3}\textrm{P}_{2}$ are the possibilities of choosing 2 of the 3 odd numbers available (3,5,7) to number the 2 faces that need to be concurrent at vertex A;
5! that is the numbers of ways we can then number the 5 left faces with the 5 numbers that are remain.
However, I can't understand why my answer is wrong. The correct answer to this problem is 1872, given with the reasoning of:
$$3!*4! + 4*3*(_{}^{3}\textrm{P}_{2})*4!=1872$$ In which:
$3!*4!$ is the numbers of ways in which all of the 4 faces that are concurrent at vertex A have odd numbers;
$4*3*(_{}^{3}\textrm{P}_{2})*4!$ is the numbers of ways in which 3 of the 4 faces that are concurrent at vertex A have odd numbers:
$4$ are the four even numbers available (2,4,6,8)
$3$ are the numbers of faces that can have the even numbers at the top of the octahedron;
$_{}^{3}\textrm{P}_{2}$ are the possibilities of choosing 2 of the 3 odd numbers available (3,5,7) to number the 2 faces that need to be concurrent at vertex A;
$4!$ that is the number of ways we can then number the 4 left faces with the 4 numbers that remain.
Even though I could understand the solution to this problem, I still don't know why mine is wrong. I am assuming I am counting extra scenarios (since $2160>1872$), but which one's?
Your formula counts the cases where all four faces are labeled with odds numbers $3$ times each, once for each pair of odd numbers $>1$. So, you need to subtract twice the number of such labelings. The offical solutions shows that there are $144$ such labelings, and indeed $$2160-288=1872$$