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If automobile license plates consist of two letters followed by four digits how many different license plates ae possible if letters and numbers can be repeated?
There are 26 possible letters and 10 possible numbers. The number of license plates possible is then 26*26*10*10*10*10 = 6760000.
Can you have two different license in two different states?
This is specificaly not allowed and can be met with fines and loss of both Licences.
What is the total number of outcomes you will have when choosing the first two characters for a license plate if it begins with a letter of the alphabet followed by a digit?
There are 26 letters (A-Z) that can be chosen for the first character, for each of which there is a choice of 10 digits (0-9) for the second character. → number of outcomes = 26 x 10 = 260.
How many 5 letters license plates are possible?
If you assume 26 different letters, there are 26 to the power 5 possible variations of 5 letters.
What is the probability of getting a license plate that has a repeated letter or digit if you live in a state that has one numeral followed by six letters followed by one numeral?
Probability of getting a repeated digit = no. of favourable outcomes/total no. of possible outcomes Favourable outcomes=(0,0),(1,1),(2,2).....(9,9) thus no. of favourable outcomes = 10 Considering that anyone of the 10 digits may apperar as the first numeral as well as d last numeral, No. of possible outcomes=10*10=100 hence probablity of a repeated digit=10/100=0.1 Probability of getting a repeated letter = no. of favourable outcomes/total no. of possible outcomes consider 6 blanks _ _ _ _ _ _ ,each filled with a letter. Thus no. of possible outcomes = 26^6 now consider that any two of these blanks have the same letter. Consider the two blanks filled with same letter as one blank. _ _ _ _ _ So that blank can be filled in 26 ways(i.e. you can have any of 26 alphabets as the repeated letter) The other 4 blanks can be filled with rest of 25 alphabets as the one that has already been used(the repeated letter) cannot be used again. We want all other letters to be differrent. So the next four blanks can be filled in 25*24*23*22ways. Thus, no. of favaourable outcomes = 26*25*24*23*22 Probability of getting a repeated letter=(26*25*24*23*22)/(26^6) =0.0255(approx) And, Total probablity of getting A repeated digit OR letter=0.1+0.0255=0.1255 i.e. 12.55% __________ Another approach, if the letters are able to be recurring throughout, which they usually are, you have a 1 in 95,428,956,661,682,176 of getting two of the same letters right next to each other. You first take 26^6, then it comes out as a 1 in 308,915,776 of getting a letter you want. Then you square that, because the outcome MUST be the same, so the chances of you getting that same letter increase by whatever the denominator is, which would be 308,915,776^2. Hopefully I'm right. xD
