Combinations: Why is this wrong? I want to find the number of different 5 digit numbers where 3 of the digits are different. Each digit can be from the set (1,2,3,4,5). Suppose I have the three digits $a,b,c$. We have to cases
$$a,b,c,a,a \text{  or   } a,b,c,a,b$$
In the first case we have $\frac{5!}{3!}$ different combinations and in the second we have $\frac{5!}{2!2!}$. Since $a,b,c$ are different integers from the set $(1,2,3,4,5)$, one will be able to take $5$ values, one will be able to take $4$ values, and one will be able to take $3$ values. Hence, the total number of combinations is $(\frac{5!}{3!}+ \frac{5!}{2!2!})\times5\times4\times3=3000$.
However, the answer to the question is given to be $1500$. Does anybody know where I have gone wrong?
It is worth noting that I have asked something very similar here, however, the comment section has gotten to large and also the answers seem to be suggestign I use a different method, but I want to know what is wrong with this.

You have two cases:  3-1-1 or 2-2-1.
Case 3-1-1:

*

*pick the three slots in which to place the repeated digit: $5\choose 3$.

*choose the duplicated digits: 10

*choose the two others together $9 \choose 2$
This gives 7200 possibilities.
Deal similarly with case 2-2-1.
