How many different three-digit number can be formed using the integers 1 to 6 if no integer occurs twice in a number?
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Correct Answer: Option C
Explanation:
To solve this, we are asked to find how many different three-digit numbers can be formed using the digits from 1 to 6, with no repetition of digits.
This is a permutation problem since the order of the digits matters and no digit can be repeated.
Steps:
1. Choosing the first digit:
- We have 6 choices (since any of the digits 1 to 6 can be used).
2. Choosing the second digit:
- After choosing the first digit, we are left with 5 digits, so we have 5 choices for the second digit.
3. Choosing the third digit:
- After choosing the second digit, we are left with 4 digits, so we have 4 choices for the third digit.
Total number of different three-digit numbers:
\[
6 \times 5 \times 4 = 120
\]
Thus, the total number of different three-digit numbers is \( \boxed{120} \).
To solve this, we are asked to find how many different three-digit numbers can be formed using the digits from 1 to 6, with no repetition of digits.
This is a permutation problem since the order of the digits matters and no digit can be repeated.
Steps:
1. Choosing the first digit:
- We have 6 choices (since any of the digits 1 to 6 can be used).
2. Choosing the second digit:
- After choosing the first digit, we are left with 5 digits, so we have 5 choices for the second digit.
3. Choosing the third digit:
- After choosing the second digit, we are left with 4 digits, so we have 4 choices for the third digit.
Total number of different three-digit numbers:
\[
6 \times 5 \times 4 = 120
\]
Thus, the total number of different three-digit numbers is \( \boxed{120} \).