If the code has to be four different digits the there are 10 x 9 x 8 x 7/ 4 x 3 x 2 ie 210 possibilites.
If repeated digits are permitted then there are 10 to the fourth ie 10000 possibilities, as each part of the code can be any of the 10 digits.
Without repeats there are 4 × 3 = 12 possible 2 digit numbers. With repeats there are 4 × 4 = 16 possible 2 digit numbers.
Assuming leading zeros are not permitted, then: If repeats are not allowed there are 30 possible numbers. If repeats are allowed there are 60 possible numbers.
There are twelve possible solutions using the rule you stated... 13,14,17,31,34,37,41,43,47,71,73 & 74
If two digit number can have a repeated digit (eg 44) then there are 20 possible numbers, otherwise there are 16 possible numbers: There are 4 single digit numbers; With repeats allowed there are a further 4 x 4 = 16 two digit numbers making a total of 20; Without repeats allowed there are a further 4 x 3 = 12 two digit numbers making a total of 16.
Total possible 4-digit numbers= 1000, 1001,...,9999 = 9000 Total with same digit numbers = 1111,2222,...,9999 = 9 9000 - 9 = 8991
Without repeats there are 4 × 3 = 12 possible 2 digit numbers. With repeats there are 4 × 4 = 16 possible 2 digit numbers.
The first digit can have 5 possible numbers, the second digit can have 4, the third 3, the fourth 2. 5
if the repetition is allowed the there is 6*6*6 possible ways = 216
Assuming leading zeros are not permitted, then: If repeats are not allowed there are 30 possible numbers. If repeats are allowed there are 60 possible numbers.
Forming three three digit numbers that use the numbers 1-9 without repeating, the highest product possible is 611,721,516. This is formed from the numbers 941, 852, and 763.
what is the least possible sum of two 4-digit numbers?what is the least possible sum of two 4-digit numbers?
There are twelve possible solutions using the rule you stated... 13,14,17,31,34,37,41,43,47,71,73 & 74
There are 720 of them. The three digit counting numbers are 100-999. All multiples of 5 have their last digit as 0 or 5. There are 9 possible numbers {1-9} for the first digit, There are 10 possible numbers {0-9} for each of the first digits, There are 8 possible numbers {1-4, 6-9} for each of the first two digits, Making 9 x 10 x 8 = 720 possible 3 digit counting numbers not multiples of 5.
If two digit number can have a repeated digit (eg 44) then there are 20 possible numbers, otherwise there are 16 possible numbers: There are 4 single digit numbers; With repeats allowed there are a further 4 x 4 = 16 two digit numbers making a total of 20; Without repeats allowed there are a further 4 x 3 = 12 two digit numbers making a total of 16.
Assuming that the first digit of the 4 digit number cannot be 0, then there are 9 possible digits for the first of the four. Also assuming that each digit does not need to be unique, then the next three digits of the four can have 10 possible for each. That results in 9x10x10x10 = 9000 possible 4 digit numbers. If, however, you can not use the same number twice in completing the 4 digit number, and the first digit cannot be 0, then the result is 9x9x8x7 = 4536 possible 4 digit numbers. If the 4 digit number can start with 0, then there are 10,000 possible 4 digit numbers. If the 4 digit number can start with 0, and you cannot use any number twice, then the result is 10x9x8x7 = 5040 possilbe 4 digit numbers.
Total possible 4-digit numbers= 1000, 1001,...,9999 = 9000 Total with same digit numbers = 1111,2222,...,9999 = 9 9000 - 9 = 8991
6*5*4*3*2*1=720 possible numbers