In the simplest case, if A is the number of values a single element of a combination can have, and N is the number of elements in the combination, then the number of possible lock combinations is AN. For example: if you have a lock with a 4-digit numerical combination, any combination has 4 elements - the digits. Each digit can have 10 values - 0 through 9. So the total number of lock combinations is 104 or 10000. Another example: if you have the typical rotating combination padlock, a combination consists of three numbers, each of which can be 0-39. So the total number of combination shere is 403 or 64000. The calculation gets more complicated if the same number can't appear twice in the combination. In that case, there are A values for the first element, but only A-1 values for the second, A-2 values for the third, and so on. The formula in this case is A(A-1)(A-2)(...)(A-N+1) which can be written more concisely as A!/(A-N)!. For example: if you have the same rotating combination padlock as above, but know that the numbers in the combination are all different, there are 40*39*38 or 59280 possible combinations.
Substitute the number in the equation. If the resulting statement is true the number is a solution to the equation.
To find the number of combinations of dimes, nickels, and pennies that equal 19 cents, we can use a systematic approach. Let's denote the number of dimes as D, nickels as N, and pennies as P. Since 1 dime is equivalent to 10 cents, 1 nickel to 5 cents, and 1 penny to 1 cent, we can set up the equation 10D + 5N + 1P = 19. To solve this equation, we can use techniques such as trial and error, substitution, or elimination to determine the possible combinations.
If repeats are allowed than an infinite number of combinations is possible.
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
There are 2^10 = 1024 combinations, including the one consisting of no items.
Substitute the number in the equation. If the resulting statement is true the number is a solution to the equation.
2^n possible combinations
35
Since a number can have infinitely many digits, there are infinitely many possible combinations.
If repeats are allowed than an infinite number of combinations is possible.
There are infinitely many numbers and so infinitely many possible combinations.
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!
128
First you try to solve it, but you will soon realize that there is only two steps. Guess and Check.
Assuming that repeated numbers are allowed, the number of possible combinations is given by 40 * 40 * 40 = 64000.If repeated numbers are not allowed, the number of possible combinations is given by 40 * 39 * 38 = 59280.
There are countless possible liquid combinations, depending on the types of liquids you are considering (water, juice, alcohol, etc.) and how many you want to mix together. The number of combinations would be exponential, as each additional liquid increases the number of possible combinations exponentially.