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
125 There are five choices for each of the three digits (since repetition is allowed). So there are 5*5*5=125 combinations.
Just six numbers... 345, 354, 435, 453, 534 & 543
There are 4*4*3*2*1 = 96 such numbers.
2,401
Oh, dude, let me break it down for you. So, you've got 5 digits to choose from for the hundreds place, then 4 left for the tens place after you've used one for the hundreds, and finally 3 left for the units place after using 2 for the hundreds and tens. Multiply those together and you get 60 possible 3-digit numbers. Easy peasy, lemon squeezy!
64 if repetition is allowed.24 if repetition is not allowed.
If repetition of digits isn't allowed, then no13-digit sequencescan be formed from only 5 digits.
125
There are 2000 such numbers.
It is 415968.
There are 2000 such numbers.
Possible solutions - using your rules are:- 11,13,17,31,33,37,71,73 &77
9 odd numbers less than 100 can be formed. They are: 3,5,7,35,37,53,57,73 and 75.
125 There are five choices for each of the three digits (since repetition is allowed). So there are 5*5*5=125 combinations.
20 of them, if repetition is not allowed.
If repetition is allowed . . . . . 343 If repetition is not allowed . . . . . 210
Just six numbers... 345, 354, 435, 453, 534 & 543