how many combinations of 2 numbers out of 10?
434189
10 10 = 100 two digit combinations. 10 10 10 = 1000 three digit combinations. There are 60 of them. Using the same argument as above there are 60 60 = 3600 two "digit" combinations and 60 60 60 = 216000 three "digit" combinations.
1387248
Said another way, there are 11,238,513 possible number combinations giving 5 choices between 1 and 69. For the second part of the Powerball, you have to pick one number from 26 possible numbers. Thus, you simply multiply your odds from the first part by 26 to get 292,201,338.
58787
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……….
1116347
Dana F Anderson, Just interested in math is all. Kinda a hobby. Assuming the digits 0 through 9 can be used, and a number can be repeated in the combination, then there are 10^3 possible combinations (1000 - from 000 to 999). If three different numbers have to be used, then 10 x 9 x 8 = 720 possible combinations.
505426
If you have a three-digit phone number, the first and second digits have 100 possible values; for each of those the third digit has 10 possible values, giving 100*10 = 1000 possible numbers. So once we get to the 10th digit, we have 10^10 possible numbers.
77076
Assuming no five-digit number can begin with zero, there are 9 possible choices for the first digit. Then there are 10 possible choices for each of the remaining four digits. Therefore, you have 9 x 10 x 10 x 10 x 10 combinations, or 9 x 10^4, which is 90,000 different combinations.
1031102
Answer and Explanation: The number of possible 16-digit combinations, where repetition is allowed is 10,000,000,000,000,000.
316456
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.
101010
The number of combinations that are possible with 10 numbers is 1,023.
219027
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!
588445
But why is this? With three columns, each with ten possibilities, the number of total combinations is 10 × 10 × 10, or 1000. With three switches, each with two possibilities, the number of total combinations is 2 × 2 × 2, or 8.
395113
[You don't want to include (40,20) and (20, 40).] So that means you need to know how many different permutations there are for each combination. If there are two numbers, there are two permutations per combination. Divide the possible permutations by number of permutations per combination: 2450 / 2 = 1225.
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4 CHOOSE 2 = 6 possible combinations. 6 is the total number of all possible combinations for choosing 2 elements at a time from 4 distinct elements without considering the order of elements in statistics & probability surveys or experiments.
22327
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.
905943
The number of combinations that are possible with 20 numbers is 1,048,575.
118583
The number of 5-digit combinations is 10 5=100,000. So, one more than 99,999. You can generalize that: the number of N-digit combinations is 10 N. Now, this assumes that you count 00000 or 00534 as "5-digit numbers".
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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.