Progetto Olimpiadi di Matematica GARA di SECONDO LIVELLO

Solution:

The answer is (B). From the fact that the number is even it follows that the last digit must be even. It is convenient to distinguish according to whether this is the doubled one or not.

In the first case the other doubled digit can occupy 3 different positions (not the 4th) and for each of these the odd digit can occupy any one of the remaining 3 positions.

In the second case the two equal digits can be arranged in 3 different ways (1 and 3, 1 and 4, 2 and 4) and for each of these the odd digit can occupy 2 positions out of the 3 free ones (not the 5th).

In total there are $3 \cdot 3 + 3 \cdot 2 = 15$ ways to fix the positions of the two equal digits and of the odd digit. For each of these ways one can freely choose the odd digit (5 ways), the doubled even digit (5 ways), the first and the second of the single even digits (4 and 3 ways respectively), for a total of $15 \cdot 5 \cdot 5 \cdot 4 \cdot 3 = 4500$ different combinations that satisfy the requirements.

Solution:

The number of ways to arrange 2 objects in $n$ spaces so that they are not consecutive is $\binom{n-1}{2}$. One can indeed think of arranging them without restrictions on $n-1$ spaces and then inserting a new empty space between them.

If we set aside for the moment the information (a), the valid ways to fix the positions of the two equal digits and of the odd digit are then $\binom{5-1}{2} \cdot 3 = 18$. Of these, the ones that do not satisfy the information (a) are $\binom{4-1}{1} \cdot 1 = 3$ (the odd digit is fixed in the last position and the doubled one has the 4 remaining spots available). In total there are then $18 - 3 = 15$ ways to fix the positions that satisfy all the information. To conclude, one multiplies by $5 \cdot 5 \cdot 4 \cdot 3$ as before.