Let us assume you can repeat numbers. That is to say, we allow 101 and are not bothered by the repeat 1, or 111 for that matter. Let's do it first so that the order matters, that is to say 123 is not the same as 321. The possible numbers for the first digit are 0,1,2,...9 so there are 10 of them, Same for the second. Now for the digit "in front", for example in 123 the digit 1, we must decide if we can allow a 0. To make things easy let's say yes for now. Then there are 10 choices for each digits and 10^3 total choices=1000 Now, lets narrow things down just in case we have some restrictions. If exclude 0 from the digit in front, then there are only 9 choices for that place and we have 900 possibilities. Now you said combinations and that usually means that order does not matter. A permutation is where order matters. So if order does not matter 123 is the same as 321 or 213. This means we over counted and now must divided. Each group of 3 numbers such as 123 can be arranged in 6 different ways, which is 3! for example 123 132 213 232 312 321 We we divided by 3! Assuming the leading 0 was allowed we have 1000/3! combinations, if not we have 900/3! combinations.
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How many four digit combinations can be made from the number nine? Example, 1+1+2+5=9.
If the same 7 digits are used for all the combinations then n! = 7! = 7*6*5*4*3*2*1 = 5040 combinations There are 9,999,999-1,000,000+1=9,000,000 7-digit numbers.
For the first digit you have 5 options, whichever you choose for the first digit, you have 4 options for the second digit, etc.; so the number of combinations is 5 x 4 x 3 x 2.For the first digit you have 5 options, whichever you choose for the first digit, you have 4 options for the second digit, etc.; so the number of combinations is 5 x 4 x 3 x 2.For the first digit you have 5 options, whichever you choose for the first digit, you have 4 options for the second digit, etc.; so the number of combinations is 5 x 4 x 3 x 2.For the first digit you have 5 options, whichever you choose for the first digit, you have 4 options for the second digit, etc.; so the number of combinations is 5 x 4 x 3 x 2.
There are 9 1-digit numbers and 16-2 digit numbers. So a 5 digit combination is obtained as:Five 1-digit numbers and no 2-digit numbers: 126 combinationsThree 1-digit numbers and one 2-digit number: 1344 combinationsOne 1-digit numbers and two 2-digit numbers: 1080 combinationsThat makes a total of 2550 combinations. This scheme does not differentiate between {13, 24, 5} and {1, 2, 3, 4, 5}. Adjusting for that would complicate the calculation considerably and reduce the number of combinations.
Just 1.