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pizza
If repetitions are allows, 325. If no repetitions are allows, then it is 32 x 31 x 30 x 29 x 28.
There are 5*4*3 = 60 such numbers.
There are 5 numbers of 1 digit, 25 numbers of 2 digits, and 75 numbers of 3 digits. This makes 105 numbers in all.
The number of possible combinations using 4 distinct numbers depends on whether the order matters and whether repetitions are allowed. If order does not matter and repetitions are not allowed, the number of combinations of 4 numbers chosen from a larger set can be calculated using the combination formula (C(n, r) = \frac{n!}{r!(n-r)!}), where (n) is the total number of numbers available. If order matters, you would use permutations instead. Please specify if you need combinations with or without repetitions and whether order matters for a more precise answer.
Infinitely many. If you restrict yourself to positive numbers only, but allow repetitions, then there are 15. If you disallow repetitions, only 2.
There are 52.
Yes, numbers are unlimited.
pizza
Considering negative numbers, the amount is unlimited.
If repetitions are allows, 325. If no repetitions are allows, then it is 32 x 31 x 30 x 29 x 28.
coolkidz
There are 5*4*3 = 60 such numbers.
There are 5 numbers of 1 digit, 25 numbers of 2 digits, and 75 numbers of 3 digits. This makes 105 numbers in all.
27 three digit numbers from the digits 3, 5, 7 including repetitions.
The Egyptians were the first people to have some knowledge in prime numbers. Though, the earliest known record are Euclid's Elements, which contain the important theorem of prime numbers. The Ancient Greeks, including Euclid, were the first people to find prime numbers. Euclid constructed the Mersenne prime to work out the infinite number of primes.
The number of possible combinations using 4 distinct numbers depends on whether the order matters and whether repetitions are allowed. If order does not matter and repetitions are not allowed, the number of combinations of 4 numbers chosen from a larger set can be calculated using the combination formula (C(n, r) = \frac{n!}{r!(n-r)!}), where (n) is the total number of numbers available. If order matters, you would use permutations instead. Please specify if you need combinations with or without repetitions and whether order matters for a more precise answer.