Questions: (a) How many three-digit numbers can be formed from the digits 0,1,2,3,4,5, and 6 if each digit can be used only once? (b) How many of these are odd numbers? (c) How many are greater than 330?

Solution Steps

(a) To form a three-digit number, the first digit cannot be 0. We need to choose 3 digits from the set {0,1,2,3,4,5,6} and arrange them. The first digit has 6 options (1-6), the second digit has 6 options (0-6 excluding the first digit), and the third digit has 5 options (excluding the first two digits).

(b) For the number to be odd, the last digit must be one of {1, 3, 5}. We need to choose the first two digits from the remaining digits and arrange them.

(c) For the number to be greater than 330, we need to consider different cases based on the first digit being 3, 4, 5, or 6.

To form a three-digit number from the digits \(\{0, 1, 2, 3, 4, 5, 6\}\) with each digit used only once, the first digit cannot be 0. We choose 3 digits from the set and arrange them. The total number of such permutations is:

\[ \text{Total permutations} = 6 \times 6 \times 5 = 180 \]

For the number to be odd, the last digit must be one of \(\{1, 3, 5\}\). We choose the first two digits from the remaining digits and arrange them. The total number of such permutations is:

\[ \text{Odd permutations} = 3 \times 6 \times 5 = 75 \]

For the number to be greater than 330, we consider different cases based on the first digit being 3, 4, 5, or 6:

• If the first digit is 3, the second digit must be greater than 3, or if the second digit is 3, the third digit must be greater than 0.
• If the first digit is 4, 5, or 6, any combination of the remaining digits will work.

The total number of such permutations is:

\[ \text{Permutations greater than 330} = 105 \]

Final Answer