In how many ways can you arrange the digits of the number $370612655$ to get a 9 digit odd number such that both $6$s come after the $2$?

Question

For example, “527650163” is acceptable, whereas “765503261” and “715503266” are not.

My attempt:

In $370612655$,

$3$, $7$, $0$, $1$, and $2$ appear once, and $6$ and $5$ appear twice.

$$\fbox{X}\fbox{X}\fbox{X}\fbox{X}\fbox{X}\fbox{X}\fbox{X}\fbox{X}\fbox{X}$$

For the last box there are $4$ options ($3$,$7$,$1$,$5$). For the first box, there are then $8$ options. Actually, $0$ cannot be placed in the first box, so there are in reality $7$ options. For the remaining $7$ boxes in the middle there are $^{7}P_7$ choices. So, there are

$$7\cdot ^{7}P_7\cdot 4$$

$$=7\cdot7!\cdot4$$

But we have not accounted for repetitions. $6$ and $5$ appear twice, so we have overcounted. Also, $6$ always has to come after $2$. We have to account for that too. $5$ appears two times, so we have to divide by $2!$, and there are $2$ $6$'s and one $2$. In total, there are $2+1=3$ if we consider that $2$ and $6$ are indistinguishable. So, we have to divide by $3!$ too. So the final answer is

$$\frac{7\cdot7!\cdot4}{2!3!}$$

Am I correct?

Answer 1

Step 1: pick the odd number at the end. You have four options, but one of those is special. Break into cases based on which it was:

Case 1: if the odd number in the units digit position was a 5

Case 2: if the odd number in the units digit position was a 1,3 or 7

Regardless which case it was, then pick the position of the $0$. It may not be the front. (I find it easier to phrase this way so we don't have to further break into cases)

Now, arrange the remaining numbers... using your idea of treating the 2 and the 6's as indistinguishable for now. Here is where which case we are in is relevant. If it was that we have only a single 5 remaining, we will not need to divide by $2$ to account for symmetry. It is only in the event that a 5 was not used as the units digit that we will have two indistinguishable 5's to concern ourselves with for symmetry.

Finish by clarifying that the 2 takes the first of the three positions selected for the 2 and 6's.

This gives an answer of:

$$1\cdot 7 \cdot \frac{7!}{3!} + 3\cdot 7\cdot \frac{7!}{2!3!} = 14700$$

Answer 2

I'm coming up with something different:

Place the $0$, There are $7$ choices for this, since $0$ can't go in the first or last slot.

Pick $3$ of the remaining slots for $2, 6, 6$. This can be done in ${7}\choose{3}$ ways--we can't use the last slot; and we can't use the slot where $0$ is.

Note that once the slots for $2,6,6$ have been chosen, there is only one way to place these three digits, since the $2$ must come first.

Pick two of the remaining five slots for the $5$s. This can be done in ${5}\choose {2}$ ways.

Arrange $1, 3, 7$ in the remaining three slots. This can be done in $3!$ ways.

Total: $7\cdot {{7}\choose{3}}\cdot {{5}\choose{2}}\cdot 3!=14,700$ ways.