# Possible Permutations for 6-Position Car Plate: Numbers and Alphabets (ABC 123)

• ZeroPivot
In summary, the problem involves finding the number of possible permutations for a car plate with 6 positions, 3 of which are numbers and 3 of which are alphabetic. The solution involves calculating the factorial of the number of options for each category and then multiplying them together. This gives a final answer of (10!/(10-3)!)*(26!/(26-3)!), which excludes permutations where the same number or alphabet is used more than once in a single plate. If repeated objects are allowed, the number of possible permutations will increase.
ZeroPivot

## Homework Statement

Lets say a car plate has 6 positions, three of them are numbers and 3 of them are alphabetic.
e.g. ABC 123. I need to figure out how many possible permutations can occur using all the numbers and alphabets BUT i cannot use any number or alfabet twice in an permutation eg. AAB 123 or ABC 112 is forbidden.

## The Attempt at a Solution

first i split the problem, first the numbers then the alfabet:

NUMBERS:
so N1 is the number of numbers i can use which is 10 obviously. 0-9
K is the number of positions which is 3.

N1!/(N1-K)=10!/(10-3)!

ALFABET:
so N2 is the number of alfabets which is 26, K is the number of positions 3.

N2!/(N2-K)!=26!/(26-3)!

The ANSWER:

(N1!/(N1-K))*(N2!/(N2-K)!) = (10!/(10-3)!)*(26!/(26-3)!)

is the answer correct?

That is correct, assuming every car plate has to use 3 letters and 3 digits. You can calculate that number as a final result.

1 person
mfb said:
That is correct, assuming every car plate has to use 3 letters and 3 digits. You can calculate that number as a final result.

thanks, does the answer exclude the permutations where the same number or digit happens in all three slots? e.g. AAA 123 or ABC 111 ?

ZeroPivot said:
thanks, does the answer exclude the permutations where the same number or digit happens in all three slots? e.g. AAA 123 or ABC 111 ?
Yes, what you have is $$3! \cdot{10 \choose 3} \times 3! \cdot {26 \choose 3}$$ which means, in the first term you select 3 distinct objects from the group of size 10. Each choice of three objects then has 6 possible rearrangements. Similar for the other term.

If you are allowed to use the same objects more than once, then clearly the number of possible permutations will increase.

1 person

## 1. What is the total number of possible permutations for a 6-position car plate with numbers and alphabets?

The total number of possible permutations for a 6-position car plate with numbers and alphabets is 26^3 * 10^3 = 17,576,000. This is because there are 26 letters in the alphabet and 10 numbers to choose from for each position.

## 2. Can the numbers and alphabets be repeated in the 6 positions?

Yes, the numbers and alphabets can be repeated in the 6 positions. This means that a car plate can have a combination such as "ABC 123" and "ABC 321".

## 3. How many possible permutations are there if the car plate can only have letters in the first 3 positions?

If the car plate can only have letters in the first 3 positions, the total number of possible permutations would be 26^3 = 17,576. This is because there are 26 letters to choose from for each of the first 3 positions, and the last 3 positions would have no restrictions.

## 4. Can the letters and numbers be in any order in the 6 positions?

Yes, the letters and numbers can be in any order in the 6 positions. This means that a car plate can have a combination such as "1A2B3C" and "C3B2A1".

## 5. How many possible permutations are there if the car plate can only have numbers in the last 3 positions?

If the car plate can only have numbers in the last 3 positions, the total number of possible permutations would be 10^3 = 1,000. This is because there are 10 numbers to choose from for each of the last 3 positions, and the first 3 positions would have no restrictions.

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