It is a permutation problem (any number can be used repeatedly), it will become 4 x 4 x 4 x 4 = 256 four digit numbers. If each number can only be used once, then the answer is 4 x 3 x 2 x 1 = 24 four digit numbers.
There are 28706 such combinations. 5456 of these comprise three 2-digit numbers, 19008 comprise two 2-digit numbers and two 1-digit numbers, 4158 comprise one 2-digit number and four 1-digit numbers and 84 comprise six 1-digit numbers.
There are 4 ways to chose the first digit. Then because there is to be no repetition, there are 3 ways to choose the second. This means that there are 4 x 3 ways of setting down the first two digits 12 _ _ 13 _ _ 14 _ _ 21 _ _ 23 _ _ 24 _ _ 31 _ _ 32 _ _ 34 _ _ 41 _ _ 42 _ _ 43 _ _ .... now you can take it from there.
Add the two greatest possible four digit numbers. 9999 + 9999
8100
7
Twelve of them.
1234, 2637, 7777, 2727 are examples of a four digit number. So any random four numbers are four digits.
4*3*2*1 = 24
256
1234
4 x 3 x 2 ie 24
1234, 1235 and 1236.
1234?
How many place combinations in a four digit number and all have to go in different spot example: 1234, 4321, 2134 etc. numbers can't change and have to stay four numbers?The answer would be 36 different combinations. This is simply the counting principle and it's easy for anyone to learn.
1234
The four digit multiples of 1234 are:1234 (1234x1)2468 (1234x2)3702 (1234x3)4936 (1234x4)6170 (1234x5)7404 (1234x6)8638 (1234x7)9872 (1234x8)Multiplying 1234 by 9 give the 5 digit number 11106, so the largest 4 digit multiple of 1234 is 9872.
Total possible 4-digit numbers= 1000, 1001,...,9999 = 9000 Total with same digit numbers = 1111,2222,...,9999 = 9 9000 - 9 = 8991