When you have a choice of 7 digits, and cannot repeat any digit, then the first number can be chosen in 7 ways, the second can be chosen in 6 and the third can be chosen in 5.
7 x 6 x 5 = 210
There are 2000 such numbers.
# 230 # 203 # 320 # 302
There are 60480 numbers.
With repetition, 94 = 6561. Without repetition, 9*8*7*6 = 3024.
You have seven different digits (symbols) to choose from, so you can form seven different one digit numbers and 7×7=72=49 different two digit numbers.
5040 different 4 digit numbers can be formed with the digits 123456789. This is assuming that no digits are repeated with each combination.
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.
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)
9*8*7 = 504 of them.
3*2*1 = 6 of them.
It is 415968.
There are 2000 such numbers.
Possible solutions - using your rules are:- 11,13,17,31,33,37,71,73 &77
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.
# 230 # 203 # 320 # 302