If repeats are allowed than an infinite number of combinations is possible.
When trying to work out how many different combinations there are, you need to know how many options there are for each value. If the password only contains lower case letters, then we have 26 options for each value. For each letter in the password, there are 26 options, so the total number of possible options is 26x26x26x26x26x26 or 266 This equals 308,915,776 so there are 308,915,776 possible different combinations of six letters.
Three combinations: 23, 24 and 34
To calculate the number of possible combinations from 10 items, you can use the formula for combinations, which is nCr = n! / r!(n-r)!. In this case, n is the total number of items (10) and r is the number of items you are choosing in each combination (which can range from 1 to 10). So, if you are considering all possible combinations (r=1 to 10), the total number of combinations would be 2^10, which is 1024.
61
Since a number can have infinitely many digits, there are infinitely many possible combinations.
The number of combinations - not to be confused with the number of permutations - is 2*21 = 42.
If the numbers can be repeated and the numbers are 0-9 then there are 1000 different combinations.
18 different combinations. When a coin is tossed twice there are four possible outcomes, (H,H), (H,T), (T,H) and (T,T) considering the order in which they appear (first or second). But if we are talking of combinations of the two individual events, then the order in which they come out is not considered. So for this case the number of combinations is three: (H,H), (H,T) and (T,T). For the case of tossing a die once there are six possible events. The number of different combinations when tossing a coin twice and a die once is: 3x6 = 18 different combinations.
7878
In a 7 segment display, the symbols can be created using a selected number of segments where each segment is treated as a different element.When 1 segment is used, the possible positions are 7because it can be any of the 7 segments (7C1=7).When 2 segments are used, the number of possible combinations are 7C2=21.When 3 segments are used, the number of possible combinations are 7C3=35When 4 segments are used, the number of possible combinations are 7C4=35When 5 segments are used, the number of possible combinations are 7C5=21When 6 segments are used, the number of possible combinations are 7C6=7When 7 segments are used, the number of possible combinations are 7C7=1Adding the combinations, 7+21+35+21+7+1=127Therefore, 127 symbols can be made using a 7 segment display!
48
2^n possible combinations
To calculate the number of ways a committee of 6 can be chosen from 5 teachers and 4 students, we use the combination formula. The total number of ways is given by 9 choose 6 (9C6), which is calculated as 9! / (6! * 3!) = 84. Therefore, there are 84 ways to form a committee of 6 from 5 teachers and 4 students if all are equally eligible.
35
If repeats are allowed than an infinite number of combinations is possible.
Allowing leading zeros, 109 or 1 billion.