how many combinations of 3 numbers can you have?
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There are, you see, 3 x 2 x 1 = 6 possible ways of arranging the three digits. Therefore in that set of 720 possibilities, each unique combination of three digits is represented 6 times.
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Originally Answered: How many combinations are there of 6 numbers in picking between 1 to 49? As long as it doesn't matter which order the numbers are picked in, you have: 49 C 6, or 49 choose 6. Enter this into google and it will tell you: 49 choose 6=13,983,816 different possible combinations of numbers……….
650668
The smallest 3 digit number I have is zero, written 000 and the largest is 999 and I have ever integer between 000 and 999. Thus I have 1,000 numbers or 1,000 possible combinations.
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There are 1 billion 9 digit numbers (000,000,000 through 999,999,999). There are 45 different combinations of two different numerals (10 x 9 divided by 2). There are 512 (2 to the 9th power) different permutations for any two numbers to be used in a 9 digit number.
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There are 6 = 3x2x1 ways to order 3 digits in a row. Thus the number of combinations of 3 of the 10 digits is 720/6 = 120 combinations.
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The number of combinations is 8,145,060. This is because there are 45!/(6!
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For each choice of the first two digits you have 10 choices for the third digit. Thus you have 10x10x10 = 1000 choices for the first three digits. Finally you have 10 choices for the fourth digit and thus there are 10x10x10x10 = 10 000 possible 4 digit combinations from 0-9.
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To find the number of permutations it is all about choices. You know the 4 digits, but you don't know the order. So this is 4 × 3 × 2 × 1 = 24, if all 4 digits are different.
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Assuming repetition is allowed, you can have 7-digit numbers from 1,000,000 to 9,999,999 which is a total of 9,000,000 7-digit numbers. These are all the possible 7-digit numbers. In general, there are 9 × 10^(n-1) possible n-digit numbers.
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However, as we have 4 slots, which need to be all filled at the same time in order to form a 4-digit number, we have to multiply the number of possibilities for each slot together, ie. 2 x 2 x 2 x 2 = = 16. Thus, 16 permutations are made possible.
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Jueteng. This is a numbers game that involves the combination of 37 numbers against 37 numbers. In some areas,the operators use 38 numbers. For 37 against 37 numbers, there are, 1,369 combinations; while for 38 against 38 numbers, there are, 1,444 combinations.
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I.e. there are 4 objects, so the total number of possible combinations that they can be arranged in is 4! = 4 x 3 x 2 x 1 = 24.
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There are about 1,000 combinations possible for a 3-digit number/code. There are 10,000 combinations possible for a 4-digit number/code. You are much better off choosing the 4-digit code lock because anyone can easily brute-force the combination in between 20 minutes and an hour.
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Again there ae 4 choices so the number of possible 3 digit numbers is 4 4 4. Finally there are 4 choices for the last digit so the number of possible 4 digit numbers is 4 4 4 = 256.
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See for 11 numbers there can be 9 ways for each number for 1st nine places since 0 can't be the fist number and next two are repeated from 10 numbers hence 10!/2! So 99 combinations have 10!/2! Permutations hence total ways are 99*10!/2!
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Thus you have made 5 × 4 × 3 × 2 1 = 120 choices and there are 120 possible 5 digit numbers made from 1, 2, 3, 4 and 5 if you don't allow any digit to be repeated.
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A standard 3-digit lock has 999 possible combinations, which is why the 4- digit locks cost a bit of a step more money, as they have 9999 possibles. One very important thing to remember is the resetting of this lock to something much different from your actual combo, when you lock it.