Assuming the digits cannot be repeated, there are 7 combinations with 1 digit, 21 combinations with 2 digits, 35 combinations with 3 digits, 35 combinations with 4 digits, 21 combinations with 5 digits, 7 combinations with 6 digits and 1 combinations with 7 digits. That makes a total of 2^7 - 1 = 127: too many for me to list. If digits can be repeated, there are infinitely many combinations.
Oh, dude, you're hitting me with some math vibes here. So, if you have 6 digits to choose from to make a 4-digit combination, you can calculate that by using the formula for permutations: 6P4, which equals 360. So, like, you can make 360 different 4-digit combinations from those 6 digits. Math is wild, man.
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If you exclude numbers starting with zero then the first digit must be between 1 and 9 (i.e. 9 combinations). The remaining 9 digits can be any value between 0 and 9 (i.e. 10 combinations).So you can have 9x109 = 9,000,000,000 combinations.
If you have 4 positions, each of which can hold any of the ten digits, you have 10 to the power 4 combinations. If you can have only 4 different digits, you have 4 to the power 4 different combinations.
Oh, dude, you're hitting me with some math vibes here. So, if you have 6 digits to choose from to make a 4-digit combination, you can calculate that by using the formula for permutations: 6P4, which equals 360. So, like, you can make 360 different 4-digit combinations from those 6 digits. Math is wild, man.
There are 167960 9 digits combinations between numbers 1 and 20.
Since a number can have infinitely many digits, there are infinitely many possible combinations.
If you use each number once, there are six combinations.
Assuming the digits cannot be repeated, there are 7 combinations with 1 digit, 21 combinations with 2 digits, 35 combinations with 3 digits, 35 combinations with 4 digits, 21 combinations with 5 digits, 7 combinations with 6 digits and 1 combinations with 7 digits. That makes a total of 2^7 - 1 = 127: too many for me to list. If digits can be repeated, there are infinitely many combinations.
There is only one combination. The order of the digits in combinations makes no difference. They are considered as being different if they are permutations, not combinations.
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There are 26 different letters that can be chosen for each letter. There are 10 different numbers that can be chosen for each number. Since each of the numbers/digits that can be chosen for each of the six "spots" are independent events, we can multiply these combinations using the multiplicative rule of probability.combinations = (# of different digits) * (# of different digits) * (# of different digits) * (# of different letters) * (# of different letters) * (# of different letters) = 10 * 10 * 10 * 26 * 26 * 26 = 103 * 263 = 1000 * 17576 = 17,576,000 different combinations.
If you exclude numbers starting with zero then the first digit must be between 1 and 9 (i.e. 9 combinations). The remaining 9 digits can be any value between 0 and 9 (i.e. 10 combinations).So you can have 9x109 = 9,000,000,000 combinations.
If you have 4 positions, each of which can hold any of the ten digits, you have 10 to the power 4 combinations. If you can have only 4 different digits, you have 4 to the power 4 different combinations.
9,000 - all the numbers between 1,000 and 9,999 inclusive. * * * * * NO. Those are PERMUTATIONS, not COMBINATIONS. Also, the question specified 4 digit combinations using 4 digits. The above answer uses 10 digits. If you start with 4 digits, you can make only 1 combination.
10 of 1 digit, 45 of 2 digits and 120 of 3 digits making 175 in all.