10,108,800 assuming that leading 0s are not allowed in the 3-digit numbers; that the licence authorities do not block some letter combinations because they are rude, and letters that might be confused with numbers (O and 0, I and 1, Z and 2, S and 5, B and 8) are all permitted.
First of all, There are 10 Different numbers, 0-9. And 26 Letters, A-Z. We want 3 of there numbers. so it is, 10P3= 720 (10!/7!) We also want 26P4 =358800 So the Answer is 358800*720=258336000. Which is 258Million 336 Thousand. Roughly. As Some letters may not be used or some numbers. I don't really know the working of license plates. Update: Assuming that you are allowing repetitions of both letters and numbers, it follows that you do not have to use (p(n,r)) to find the permutation. The correct calculation is: 10^3(26^4) or p= 10*10*10(26*26*26*26)= 456,976,000
If it alternates and you start with a letter, then there are 11,232,000 permutations. Then if you start with a number and alternate, there are another 11,232,000 permutations, for a total of 22,464,000 permutations. If you exclude the letters I, S, B, and O (because they look kind-of like 1, 5, 8, & 0 - kind-of important on license plates) then you are down to 6,652,800 & 13,305,600 respectively.
each of the four numbers have 10 possibilities, and each of the three letters have 26 possibilities. so the total possible ways u can arrange them are 10x10x10x10x26x26x26 this totals up to 175,760,000 different license plate numbers
When you spell out the words for numbers 1-10 four of the numbers have three letters. (one two six ten)
There are 676,000 ways to make the license plates.
There are 17,576 possible license plates. There are 26 possible letters in the first space times 26 in the second space times 26 in the third space.
263 x 98 (assuming "00" is not used) ie 1,722,448
35,152,000 (assuming that 000 is a valid number, and that no letter combinations are disallowed for offensive connotations.) Also, no letters are disallowed because of possible confusion between letters and numbers eg 0 and O.
Any three letters, in any order, including repeated letters gives 263 combinations each of which could have one of 9 digits so 26 x 26 x 26 x 9 ie 158184 different plates.
Multiply the possibilities for each digit: 26 * 26 * 26 * 10 * 10 * 10 = 17,576,000
10,108,800 assuming that leading 0s are not allowed in the 3-digit numbers; that the licence authorities do not block some letter combinations because they are rude, and letters that might be confused with numbers (O and 0, I and 1, Z and 2, S and 5, B and 8) are all permitted.
In California, for example, the first digit of a standard plate is a number, followed by 3 letters, and then three numbers. There are 26 letters in the alphabet, so there are 26 raised to the 3rd power combinations, or 26 * 26 * 26, which is 17,576 possibilities just of the 3 letters.
First of all, There are 10 Different numbers, 0-9. And 26 Letters, A-Z. We want 3 of there numbers. so it is, 10P3= 720 (10!/7!) We also want 26P4 =358800 So the Answer is 358800*720=258336000. Which is 258Million 336 Thousand. Roughly. As Some letters may not be used or some numbers. I don't really know the working of license plates. Update: Assuming that you are allowing repetitions of both letters and numbers, it follows that you do not have to use (p(n,r)) to find the permutation. The correct calculation is: 10^3(26^4) or p= 10*10*10(26*26*26*26)= 456,976,000
If it alternates and you start with a letter, then there are 11,232,000 permutations. Then if you start with a number and alternate, there are another 11,232,000 permutations, for a total of 22,464,000 permutations. If you exclude the letters I, S, B, and O (because they look kind-of like 1, 5, 8, & 0 - kind-of important on license plates) then you are down to 6,652,800 & 13,305,600 respectively.
each of the four numbers have 10 possibilities, and each of the three letters have 26 possibilities. so the total possible ways u can arrange them are 10x10x10x10x26x26x26 this totals up to 175,760,000 different license plate numbers
all numbers with three and four in it