Dickson Kataza
Feb 6
How many three digit numbers divisible by three can be formed using the digits 2, 3, 4, 5 and 6 if repetition of digits is not allowed?
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Feb 6
To form a three-digit number divisible by three, the sum of its digits must be divisible by three.
The possible combinations of three digits from the given set are:
- 2, 3, 4
- 2, 3, 5
- 2, 3, 6
- 2, 4, 5
- 2, 4, 6
- 2, 5, 6
- 3, 4, 5
- 3, 4, 6
- 3, 5, 6
- 4, 5, 6
For each combination, we can arrange the digits in $3! = 6$ ways.
Therefore, there are $10 \times 6 = \boxed{60}$ three-digit numbers divisible by three that can be formed using the digits 2, 3, 4, 5, and 6 without repetition.
The possible combinations of three digits from the given set are:
- 2, 3, 4
- 2, 3, 5
- 2, 3, 6
- 2, 4, 5
- 2, 4, 6
- 2, 5, 6
- 3, 4, 5
- 3, 4, 6
- 3, 5, 6
- 4, 5, 6
For each combination, we can arrange the digits in $3! = 6$ ways.
Therefore, there are $10 \times 6 = \boxed{60}$ three-digit numbers divisible by three that can be formed using the digits 2, 3, 4, 5, and 6 without repetition.
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