32C3 = 4960
9,752 9,725 9,572
-4309278
9*9*9 = 729 using the digits 1 to 9 and 2*9 using 10 and another digit. 749 in all.
about 13
depends on your answer
32C3 = 4960
9,752 9,725 9,572
-4309278
9*9*9 = 729 using the digits 1 to 9 and 2*9 using 10 and another digit. 749 in all.
Assuming you mean permutations of three digits, then the set of numbers that can be made with these digits is: 345 354 435 453 534 543 There are six possible permutations of three numbers.
6
24 three digit numbers if repetition of digits is not allowed. 4P3 = 24.If repetition of digits is allowed then we have:For 3 repetitions, 4 three digit numbers.For 2 repetitions, 36 three digit numbers.So we have a total of 64 three digit numbers if repetition of digits is allowed.
about 13
1
There are 4 options for the hundreds digit (3, 4, 5, or 6), and 4 options for the tens digit (including the possibility of repeating the hundreds digit). Similarly, there are 4 options for the units digit. Therefore, the total number of 3-digit numbers that can be formed using the digits 3, 4, 5, and 6 with repetition allowed is 4 x 4 x 4 = 64.
2