How many odd numbers less than $1000$ can be formed by using the digits $0,3,5,7$. Repetition not allowed.

Question

Q. How many odd numbers less than $1000$ can be formed by using the digits $0,3,5,7$. Repetition not allowed.

A. $21$

Answer is correct (please provide a thorough explanation).

Unit digit nos. : $3$

Dual digit nos. : $2×3$

Three digit nos. : $2×2×3$ (as per the answer)

Three digit nos. : $1×3×3$ (as per my viewpoint)

Please help.

Could you explain how you obtained $1 \cdot 3 \cdot 3$ for the three-digit number case? — N. F. Taussig, Jun 15 '15 at 10:34
The question is unclear. Can the number have a leading zero? — Rory Daulton, Jun 15 '15 at 10:35
@user147163 Add the number of one-digit, two-digit, and three-digit numbers. — N. F. Taussig, Jun 15 '15 at 10:36
You have three choices for the units digit $(3, 5, 7)$. You can't have $0$ because the number would be even then. Next, you have two choices for the hundreds digit. It can't be zero because then the number wouldn't be three digits, and it can't be the number chosen for the units digit. Finally, there are two remaining choices for the tens digit. The challenge in the calculation is that when considering the digits in order, a zero or not in the tens place requires two cases. — Michael Burr, Jun 15 '15 at 10:42

score 1 · Accepted Answer · answered Jun 15 '15 at 10:38

1

For the three-digit number case, there are three ways of selecting the units digit since we cannot use $0$. Once the units digit has been chosen, there are two choices for the hundreds digit since we cannot use either zero or the units digit. That leaves us with two choices for the tens digit since we cannot use the hundreds digit or the units digit. Hence, the number of three-digit odd numbers that can be formed using only the digits $0, 3, 5, 7$ is $2 \cdot 2 \cdot 3$.

answered Jun 15 '15 at 10:38

N. F. Taussig

76,571

Plz tell why the 10th digit no. can't be 0? – Kountay Dwivedi Jun 15 '15 at 11:10
I mean, if I take the unit as 3choices, tens as 3choices including 0; then that leaves hundreds as 1choice. That is: 1,3,3 – Kountay Dwivedi Jun 15 '15 at 11:13
The tens digit can be zero. The two digits that cannot be used in the tens place are the hundreds digit and the units digit, both of which are odd numbers. That leaves us with two choices for the tens digit, the remaining odd number and zero. – N. F. Taussig Jun 15 '15 at 11:13
Aaaaaaah ! Now I got it. Thank you. Thank you so much. – Kountay Dwivedi Jun 15 '15 at 11:16
You're welcome. – N. F. Taussig Jun 15 '15 at 11:16

score 1 · Answer 2 · edited Jun 12 '20 at 10:38

As I understand it, you think like the following: number xyz, 3 choices for x (3,5 or 7), then there are 3 choises for y (the two remaining odd numbers and 0) and then for choosing z there is only one choise, since it needs to be odd. Thus we have $3\cdot 3\cdot 1$ different numbers.

This is not correct since if $y=0$ then we have two different choises for the number $z$ not 1. Thus we may do callculations like this:

If $y=0$ then we have $3\cdot 1 \cdot 2$ choises.

If $y\neq 0$ then we have $3\cdot 2\cdot 1$ choises.

Thus in total we have $3\cdot 2\cdot 1 + 3\cdot 1\cdot 2 = 3 \cdot 2 \cdot 2$ choises.

How many odd numbers less than $1000$ can be formed by using the digits $0,3,5,7$. Repetition not allowed.

2 Answers2