120.
There are 6 digits. If we pick the digits in order, there are
6 possible digits for the first digit
5 remaining digits for the second digit
4 remaining digits for the third digit.
6*5*4 = 120.
As there are 5 different digits, allowing for repetition of a digit, the first digit can be any of the 5, and for each of these, the second can be any of the 5, and so for the third, making: 5 x 5 x 5 = 125 different 3-digit numbers.
-4
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.
125 There are five choices for each of the three digits (since repetition is allowed). So there are 5*5*5=125 combinations.
6 x 6 x 6 x 3 = 648 6 because the first digit can be any of the numbers 6 again, because the second digit can be any of the numbers 6, again, because the third digit can be any of the numbers 3, because the fourth/last digit can only be 2, 4, or 6
I take it that you want to make three digits numbers with 8,7,3, and 6 without repetition. The first digit cane be selected from among 4 digits, the second from 3 digits, the third digit from 2, hence the number of three digit numbers that can be formed without repetition is 4 x 3 x 2 = 24
Six (6)
There are 600 such numbers.
There are 504 such numbers.
If repetition is allowed . . . . . 343 If repetition is not allowed . . . . . 210
9*8*7 = 504 of them.
3*2*1 = 6 of them.
64 if repetition is allowed.24 if repetition is not allowed.
9999 - 102 = 9897
# 230 # 203 # 320 # 302
1
9999-1023 = 8976