how many 4 number combinations can there be from 0-99 without repeating a number?
100 choose 4, or C(100,4) where
C(n,r) is defined as
C(n,r)=n!/[r!(n-r)!]
Think of choosing the first number out of 100 choices. We can choose the next number from the remaining 99 numbers, then 98, then 97.
By the rule of multiplication of a multi-step experiment, we have
100*99*98*97 possible outcomes, which correspond exactly to C(100,4).
To calculate the number of 4-digit combinations without repeating a number from 0 to 99, you can use the concept of permutations.
Since there are 100 numbers from 0 to 99, the first digit can be chosen from 100 possibilities.
After selecting the first digit, the second digit can be chosen from the remaining 99 possibilities (since you cannot repeat the first digit).
Similarly, the third digit can be chosen from the remaining 98 possibilities.
Finally, the fourth digit can be chosen from the remaining 97 possibilities.
Therefore, the total number of 4-digit combinations without repeating a number is calculated as follows:
100 * 99 * 98 * 97 = 941,094,000 combinations.
So, there are 941,094,000 different 4-digit combinations possible without repeating a number from 0 to 99.
To find the number of 4-number combinations without repeating a number from 0 to 99, we can use the concept of permutations.
In this case, we need to choose 4 numbers out of a set of 100 numbers (0 to 99). We cannot repeat numbers, so each number chosen must be unique.
The formula to calculate permutations is given by nPr = (n!)/((n-r)!), where n is the total number of items in the set and r is the number of items we want to choose.
In our case, n = 100 (total numbers in the set) and r = 4 (number of numbers to be chosen). Therefore, we can calculate the number of combinations as:
100P4 = (100!)/((100-4)!)
= 100!/(96!)
= (100 * 99 * 98 * 97 * 96!)/(96!) [Simplifying the factorial terms]
= 100 * 99 * 98 * 97
= 3,921,600
So, there are 3,921,600 possible 4-number combinations without repeating a number from 0 to 99.