Any admissible string of five digits uses one of the digits $1$, $2$, $3$, $6$ exactly twice and the remaining digits exactly once. There are $4\cdot{5!\over 2!}=240$ such strings, since you can choose the digit appearing twice in four ways and then arrange the resulting quintuple in ${5!\over2!}$ ways. Exactly half of these strings have an even digit at the end. It follows that there are $120$ numbers of the required kind.
Your answer is different because in the question, repitition is allowed, but you have only chosen combinations $(1,2,3),(1,2,4),(1,3,4),(2,3,4)$ in which there are no repeat numbers. So the combinations that you were supposed to include were $(1,1,1),(2,2,2),(3,3,3),(4,4,4), (1,2,2), (1,3,3),(1,4,4)...$ and so on.
Now, if you count the permutations for each of these combinations separately, and add it to 24, you will get 64.
Hope this helps :)
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Best Answer
Just to nitpick the original problem, I see only two numbers: $1$ and $2.$ It seems you are meant to write each one three times. Moreover, I suppose you are meant to leave three of the places blank.
So you are asked to make an arrangement in the nine places such that three places are filled with $1$, three are filled with $2$, and three are left blank.
Now if the problem were (for example) merely to fill three places with $1$ and leave the other six blank, you would use the binomial coefficient $\binom 93$ or the binomial coefficient $\binom 96$, depending on whether you are thinking more about the three filled places or the six blank places.
Binomial coefficients are good if there are only choices for each place, such as put $1$ in it or leave it blank. When there are three or more choices, you can use multinomial coefficients.
Since you want three of each of the three kinds of "filling" ($1,$ $2,$ or blank) in your nine places, the multinomial coefficient that counts the number of ways to fill the places is
$$ \binom{9}{3,3,3} = \frac{9!}{3!3!3!}.$$
Note that you can derive the multinomial coefficient formula by seeing that you have $\binom 93$ ways to choose which places are filled with $1,$ and then for each of those ways to fill places with $1$ you have $\binom 63$ ways to choose what is written in the remaining places. So $$ \binom{9}{3,3,3} = \binom 93 \binom 63.$$