How many four-digit numbers divisible by 10 can be formed using 1, 5, 0, 6, 7 without repetition of digits?
24
We are asked to find the number of four-digit numbers that can be formed using the digits 1, 5, 0, 6, and 7 without repeating any digit. Additionally, the numbers must be divisible by 10.
A key constraint is that the four-digit number must be divisible by 10. A number is divisible by 10 if and only if its units digit is 0.
The available digits are 1, 5, 0, 6, and 7. Since 0 is available in this set, we can form numbers divisible by 10 by placing 0 in the units position.
The digits used to form the four-digit number must not be repeated. This means once a digit is used in one position, it cannot be used in any other position.
Let the four-digit number be represented by four positions: Thousands, Hundreds, Tens, and Units.
_ _ _ _ (Thousands, Hundreds, Tens, Units)
To find the total number of such four-digit numbers, we multiply the number of choices for each position, following the fundamental principle of counting.
Total number of four-digit numbers = (Choices for Thousands) $\times$ (Choices for Hundreds) $\times$ (Choices for Tens) $\times$ (Choices for Units)
\text{Total numbers} = 4 \times 3 \times 2 \times 1
\text{Total numbers} = 24
Therefore, 24 distinct four-digit numbers divisible by 10 can be formed using the digits 1, 5, 0, 6, 7 without repetition.
This table summarizes the choices for each digit position when forming the four-digit number divisible by 10 without repetition.
| Position | Constraint | Available Digits (from 1, 5, 0, 6, 7) | Number of Choices |
|---|---|---|---|
| Units | Must be 0 for divisibility by 10 | {0} | 1 |
| Thousands | Cannot be 0 (used in Units), no repetition | {1, 5, 6, 7} (remaining after fixing Units) | 4 |
| Hundreds | No repetition (2 digits used already) | 3 digits remaining | 3 |
| Tens | No repetition (3 digits used already) | 2 digits remaining | 2 |
This problem is an application of permutations and the fundamental principle of counting.
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