Question

1. license plates a certain states license plate has 3 letters followed by 4 numbers. repeats are not al - lowed for the letters, but they are for the numbers. how many such license plates are possible? if they are issued at random, what is the probability that the 3 letters are 3 consecutive letters in alphabetical order?

Explanation:

Step1: Calculate the number of possible license plates

There are 26 letters in the alphabet and 10 digits (0 - 9).
For the 3 - letter part (no repeats), the number of permutations of 26 letters taken 3 at a time is \(P(26,3)=\frac{26!}{(26 - 3)!}=26\times25\times24\).
For the 4 - number part (no repeats), the number of permutations of 10 digits taken 4 at a time is \(P(10,4)=\frac{10!}{(10 - 4)!}=10\times9\times8\times7\).
The total number of license plates \(N = P(26,3)\times P(10,4)=26\times25\times24\times10\times9\times8\times7=78624000\).

Step2: Calculate the number of license plates with 3 consecutive letters in alphabetical order

Let the three - letter combination be considered. If the three letters are in alphabetical order, for any three distinct letters \(a,b,c\) with \(a\lt b\lt c\), there is exactly 1 way to arrange them in alphabetical order. The number of ways to choose 3 distinct letters from 26 is \(C(26,3)=\frac{26!}{3!(26 - 3)!}=\frac{26\times25\times24}{6}\). And the number of ways to arrange the 4 non - repeating numbers is still \(P(10,4)=10\times9\times8\times7\).
The number of license plates with 3 consecutive letters in alphabetical order \(M = C(26,3)\times P(10,4)=\frac{26\times25\times24}{6}\times10\times9\times8\times7 = 13104000\).

Step3: Calculate the probability

The probability \(P=\frac{M}{N}\). Since \(N = 26\times25\times24\times10\times9\times8\times7\) and \(M=\frac{26\times25\times24}{6}\times10\times9\times8\times7\), then \(P=\frac{1}{6}\).

Answer:

The number of license plates is \(78624000\). The probability that the 3 letters are in alphabetical order is \(\frac{1}{6}\).