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If the license plate code consists of (letter-1)(letter-2)(digit-1)(digit-2)(digit-3), and the letters may not repeat, and the digits may repeat, then: There are 26 choices for (letter-1, and for each of those ... there are 25 choices for (letter-2), and for each of those ... there are 10 choices for (digit-1), and for each of those ... there are 10 choices for (digit-2), and for each of those ... there are 10 choices for (digit-3). The total number of choices is: (26 x 25 x 10 x 10 x 10) = 650 x 1,000 = 650,000
The two additional digits can have the values {0, 2, 3, 4, 5, 6, 7, 8, 9} without repeats. There are 9*8 = 72 possibilities.
the word "caring" has 6 letters and none repeat so it is 6 factorial 6! = 6x5x4x3x2x1 = 720 ways
Because there are 7 letters that don't repeat, 7X6X5X4=840 combinations. To give you an idea of how to solve the problem, if there had been 9 letters and you wanted to make combinations of 3 letters, you would multiply: 9X8X7=504
If they can repeat, then: 17^6=24,137,569 If they can't repeat, then: 17*16*15*14*13*12=8,910,720
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.
Say you have the letters A,B, and C. Here are all the possible combinations. * ABC * ACB * BAC * BCA * CAB * CBA So, 6 if you don't repeat any of the letters. If you DO repeat letters, then simply take the number of letters you have, (3 for instance), and multiply it to the power of the number of letters you have. So, for 3 letters, the formula would be 33 . Or if you had 4 letters it would be 44 and so on.
Another word for "repeat" with 7 letters is:iteraterestatereprise
hi
If the license plate code consists of (letter-1)(letter-2)(digit-1)(digit-2)(digit-3), and the letters may not repeat, and the digits may repeat, then: There are 26 choices for (letter-1, and for each of those ... there are 25 choices for (letter-2), and for each of those ... there are 10 choices for (digit-1), and for each of those ... there are 10 choices for (digit-2), and for each of those ... there are 10 choices for (digit-3). The total number of choices is: (26 x 25 x 10 x 10 x 10) = 650 x 1,000 = 650,000
buckingham
to repeat letters or other documents
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If you can repeat letters, you can make the word 'eleven'.
The first number can be one of 9, the second one of 8 (after the first has been eliminated), the third one of 7. The first letter (4th digit) can be one of 25, the second one of 24, the third one of 23. Multiplying those 6 numbers gives 6,955,200 which is the number of possibilities.
It is indeed possible to repeat a grade twice. You will repeat the same grade until you improve.