Number of ways of forming $8$-digit odd number using only the digits $0, 0, 2, 2, 3, 3, 4, 5$

Question

Let $\lambda$ be the number of all possible $8$ digit odd numbers

formed by using only digits $0,0,2,2,3,3,4,5,$ Then $\dfrac{\lambda}{900}$ is

My Try: Number is odd if last digit (unit position ) is odd.

So total number of ways of choosing the units place is $\dfrac{3!}{2!}=3$

Now arranging the extreme left position is $\dfrac{6!}{2!\cdot 2!}$

and arranging all digits is $\dfrac{6!}{2!}$ ways.

But it have seems that I have done the problem incorrectly.

Could some help me to solve it? Thanks

Answer 1

Why do you have factorials when choosing the units digit and leading digit? There is no arrangement happening.

Furthermore you will have to split into a couple cases. You might try to count the number of ways to choose the units digit (it's either $3$ or $5$, so $2$ choices), then count the number of ways to choose the leading digit (it's anything but $0$, so $4$ choices). However, some of these choices are incompatible. E.g., if I choose $5$ for the units digit, it cannot be the leading digit (but choosing $3$ for the units digit and the leading digit is possible).

Beyond this, the number of ways to arrange the remaining digits depends on the digits chosen in the first two steps. For instance, if I choose $4******5$, then I have to arrange $0,0,2,2,3,3$, but if I choose $2******3$ then I have to arrange $0,0,2,3,4,5$. The number of arrangements are different in these two cases.

Answer 2

Strategy:

Since we wish to form eight-digit odd numbers, the two zeros may not be placed in the first or last positions. Choose two of the middle six positions for the zeros.

Since the number must be odd, its last digit must be a $3$ or $5$.

If the units digit is a $3$, then the remaining five positions must be filled with two $2$s, one $3$, one $4$, and one $5$. Choose two of the five positions for the $2$s. Arrange the three remaining distinct letters in the remaining three positions.

If the units digit is a $5$, then the five remaining positions must be filled with two $2$s, two $3$s, and one $4$. Choose two of the five positions for the $2$s. Choose two of the remaining three positions for the $3$s. The $4$ must be placed in the remaining position.

The number of possible eight-digit odd numbers that can be formed with the given digits is $$\binom{6}{2}\binom{5}{2}3! + \binom{6}{2}\binom{5}{2}\binom{3}{2}$$

Answer 3

Another method.

Case 1 (last digit $3$). There are $\frac{7!}{2!2!}=1260$ numbers in total, in particular, with the first digit $0$. There are $\frac{6!}{2!}=360$ numbers with the first $0$ and last $3$. Hence: $1260-360=900$.

Case 2 (last digit $5$). There are $\frac{7!}{2!2!2!}=630$ numbers, in particular, with the first digit $0$. There are $\frac{6!}{2!2!}=180$ numbers with the first $0$ and last $5$. Hence: $630-180=450$.

Adding the two cases: $900+450=1350$.