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
Best Answer
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: $900+450=1350$.