Answer:
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.
