In another state, all license plates consist of from four to six symbols chosen from the 26 letters of the alphabet together with the ten digits 0-9.

a. how many license plates are possible if repetition of symbols is allowed?

b. how many license plates do not contain any repeated symbol? h

c. how many license plates have at least one repeated symbol?

d. what is the probability that a license plate chosen at random has a repeated symbol?

Question

Callie Wolfe

Mathematics

30 April, 21:20

In another state, all license plates consist of from four to six symbols chosen from the 26 letters of the alphabet together with the ten digits 0-9.

a. how many license plates are possible if repetition of symbols is allowed?

b. how many license plates do not contain any repeated symbol? h

c. how many license plates have at least one repeated symbol?

d. what is the probability that a license plate chosen at random has a repeated symbol?

+5

Answers (1)

Know the Answer?

Sidney Hickman · Answer 1

A)

First note that the plates can have between 4 and 6 symbols so we will need to find the number of plates with 4 symbols, 5 symbols and 6 symbols. We add these to get the total. In this part repetition of symbols is allowed. Since there are 26 + 10 = 36 possible symbols we look at each position on the plate and think of how many choices there are. We multiply the number f choices using the counting principal since the choices are each independent - - one symbol does not affect another. There are 36 choices for the first symbol, 35 for the second and so on. The number of plates is:

4-symbols = (36) (36) (36) (36) = 36^4

5 symbols = (36) ^5

6 symbols = 36^6

So the total here is: 36^4+36^5+36^6

B) Here we do not repeat symbols so there are 36 choices for the first symbol but only 35 for the next and 34 for the one after and so on.

4-symbols = (36) (35) (34) (33)

5 symbols = (36) (35) (34) (33) (32)

6 symbols = (36) (35) (34) (33) (32) (31)

So the total here is: (36) (35) (34) (33) + (36) (35) (34) (33) (32) + (36) (35) (34) (33) (32) (31)

c)

In order for there to be a repeated symbol we have 36 choices for the first symbol, 36 for the next and so on. However, for the last symbol we have to pick from one of the ones already selected so there are 3, 4 or 5 choices respectively.

4-symbols = (36) (36) (36) (3)

5 symbols = (36) (36) (36) (36) (4)

6 symbols = (36) (36) (36) (36) (36) (5)

So the total here is: (36^3) (3) + (36^4) (4) + (36^5) (5)

D)

The probability is given by (the number of plates with at least one repeated symbol) / (the total number of plates if repetitions are allowed) = (the answer to c) / (the answer to a)