It depends on the number of letters in the set - which is not specified. If there are n different letters in the set, the answer is n*(n-1)*(n-2)*(n-3)*(n-4)
3 decimal digits without repeats can form (10 x 9 x 8) = 720 distinct displays.For each of these . . .3 letters without repeats can form (26 x 25 x 24) = 15,600 distinct displays.Combine them on one plate, and there are (720) x (15,600) = 11,232,000 distinct displays available.
Tandemly arranged repeats are the repetitious nucleotide sequence that occurs between genes.
Well if repeats are allowed, then unlimited. If you can repeat the numbers then the list would go on forever.
This depends on if you want repeats or not (or the state you are in) with repeats it is: 26x26x26x10x10= 1,757,600 without repeats it is: 26x25x24x10x9=1,560,000
Assuming repeats are allowed and case (of the letters) is ignored then there are: 26^2 × 10^4 = 6,760,000
Assuming leading zeros are not permitted, then: If repeats are not allowed there are 30 possible numbers. If repeats are allowed there are 60 possible numbers.
Without repeats, 24. With repeats, 256.
If repeats are allowed than an infinite number of combinations is possible.
Assuming you are using combinations in the colloquial way (which is the mathematical "permutations" where order of selection does matter) to create a 3 digit number that does not start with 0, ie creating a number that is between 100 and 999 inclusive then: If repeats are not allowed there are 3 × 3 × 2 = 18 possible numbers If repeats are allowed, then there are 3 × 4 × 4 = 48 possible numbers. If you are using combinations in the mathematical sense where order of selection does not matter and are creating groups of 3 digits, then: If repeats are not allowed there are 4 possible groups If repeats are allowed there are 20 possible groups.
It is impossible to know without more information. Huw many letters in the plate? How many times is Q allowed, IE can it be repeated? Lets say there are 3 numbers and 3 letters on the plate. If all letters are used in all 3 spaces with repeats allowed tbut maybe the beginning number cannot be zero, then 1x25x25x9x10x10 + 25x1x25x9x10x10 + 25x25x1x9x10x10 + 1x1x25x900 +1x25x1x900 + 25x1x1x900 + 1x1x1x900 = 562500x3 + 22500x3 + 900 = 1,755,900 possible plates with at least 1 Q
Irrational numbers can not be repeating decimals. Any number that is a repeating decimal is rational.