Q‒1. [5×4 marks] a) How many three-digit numbers can be formed from the digits 0, 1, 2, 3, 4, 5, and 6? (150) b) 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? c) How many odd numbers can be formed from the digits 0, 1, 2, 3, 4, 5, and 6 if each digit can be used only once? d) How many three-digit numbers greater than 330 can be formed from the digits 0, 1, 2, 3, 4, 5, and 6? e) How many three-digit numbers greater than 330 can be formed from the digits 0, 1, 2, 3, 4, 5, and 6 if each digit can be used only once?

a) 294

b) 180

c) 75

d) 174

e) 105

Step-by-step explanation:

I assume that for each problem, the first digit can't be 0.

a) There are 6 digits that can be first, 7 digits that can be second, and 7 digits that can be third.

6×7×7 = 294

b) This time, no digit can be used twice, so there are 6 digits that can be first, 6 digits that can be second, and 5 digits that can be third.

6×6×5 = 180

c) Again, each digit can only be used once, but this time, the last digit must be odd.

If only the last digit is odd, there are 3×3×3 = 27 possible numbers.

If the first and last digits are odd, there are 3×4×2 = 24 possible numbers.

If the second and last digits are odd, there are 3×3×2 = 18 possible numbers.

If all three digits are odd, there are 3×2×1 = 6 possible numbers.

The total is 27 + 24 + 18 + 6 = 75.

d) If the first digit is 3, and the second digit is 3, there are 1×1×6 = 6 possible numbers.

If the first digit is 3, and the second digit is greater than 3, there are 1×3×7 = 21 possible numbers.

If the first digit is greater than 3, there are 3×7×7 = 147 numbers.

The total is 6 + 21 + 147 = 174.

e) If the first digit is 3, and the second digit is greater than 3, then there are 1×3×5 = 15 possible numbers.

If the second digit is greater than 3, there are 3×6×5 = 90 possible numbers.

The total is 15 + 90 = 105.

they can both be solutions depends on the equation

step-by-step explanation:

what is the equation? ?