Given the following letters in a license plate, how many combinations of them can you create
AAAA1234
Please note that this is not a homework question (I am too old for college :)
I am only trying to understand permutations and combinations. I always get lost when I see questions like this. Do I use n! or nPr or nCr.
Any book on this subject in addition to the logic used to arrive at the answer will also be greatly appreciated.
How many distinct sets of positions can the A's occupy? Given this value, multiply by the number of distinct arrangements of 1234 and you have your answer. You'll need to choose the positions for the A's and then ! will help with the arrangements of 1234.
If you mean 3 letters A-Z and 4 digits 0...9 in that order, then you have
26 letters x 26 letters x 26 letters x 26 letters x 10 digits x 10 digits x 10 digits x 10 digits
= 26^4 * 10^4
= 4569760000
If no leading "0" is allowed, you get a few less.
Edit1: Miscounted the "A"
Edit2: I reread the question - originally I thought it was just four letters at the beginning followed by 4 numbers. If it's just a permutation thing, then the answer is obviously different: 8! permutations at all, but 4! permutations for the A are the same, so 8! / 4! = 1680.
One strategy that you can use (there will be many) is to get all the permutations possible, then divide out the repeats.
Permutations of 8 elements = 8!
But for each unique arrangement of these, there are a bunch more with the same positions of the A's. So, how many ways can you arrange four A's in one particular set of positions?
Permutations of 4 A's = 4!
So the total unique arrangements should be 8! / 4!
If I'm totally wrong just someone say so and I'll delete this answer...
Consider a simpler example. Let's say you had asked the question:
How many arrangements are there of the symbols: ABCD1234? Now, since every symbol is distinct, there are 8! ways to arrange them.
Now let's build up to your problem. If we change the letter B to an A, we have AACD1234. This destroys the uniqueness of exactly half the possible combinations, since any combination where we could have previously switched the A and the B is now non-unique. Therefore, we now have 8!/2 combinations.
Similarly, replacing the C with another A would result in half of the remaining combinations losing their uniqueness, and so on.
So, if only one symbol is duplicated, the generalized formula is (number of symbols total)!/2^(number of duplications)
In your case, the number of possible arrangements is 8!/2^4
I have faith in exactly one method to remember such formulas: Rethink through the reasoning to justify it as needed. Then, each time you need the formula, remembering it becomes a mental exercise that makes it easier to remember it the next time. It also allows you to know the math on your own authority, instead of someone else's authority.
If the letters are all different, then there are n choices for the first letter, n-1 choices for the second letter, and so on. That makes n! However, in your problem the letters are not all different. One trick is to tag them to make them different so that you are overcounting, then divide by the amount that you are overcounting. If a of the symbols are A, then you can tag them in a! ways. They are then all different, so that the answer to the modified question is n!. So the answer to the original question is n!/a! (This is assuming that the symbols other than the A are fixed, distinct numbers.)
Another argument is to count the positions for the numbers. There are n positions for the 1, n-1 positions for the 2, etc., so you get n(n-1)...(n-r+1) = n!/a!, where r = n-a.
In fact the answer is the same as the permutation formula nPr. And your arrangements are much the same as partial permutations, which is what the formula is for. But you'll learn it better if you reason through it before looking at the formula.
As for books, I might suggest Brualdi, Introductory Combinatorics.
Answer is 8!/4!
Let's try to explain with a simpler question: Combinations of 112 ? There are 112, 121 and 211. If all digits would be unique, we could just find the answer by 3! But there is a repeating digit. So we should extract repeating digits by 3!/2! = 3
Another example is 1122. We have two repeating digit here. So we should extract twice. 4!/2!.2! = 6
No need for permutations, because all letters can be repeated, even the number
since the given example is [AAAA1234],then we have 4-Letters and 4-Digits.
for each letter we have 26 {A-Z} possible combinations
Thats why for 4 letters we will have 26^4
For each Number we have 10 {0-9} possible combinations, except the last digit we 9 possible combinations {case 1}, if it not allowed to be 0 otherwise it is 10 {case 2}
Thats why for 4 letters we will have 9*10^3 {case 1} or 10^4 {case 2}
The total number of combinations is {case 1} 9*(26^4)*(10^3) or {case 2} (26^4)*(10^4)
But if your question about permutations for the set{A,A,A,A,1,2,3,4}, then consider the the equivalent set {1,2,3,4,5,6,7,8} and try avoid the repeated sequence by divide over the permutations of {5,6,7,8} and the answer is 8!/4!=5*6*7*8=1680. the{5,6,7,8} represent {A,A,A,A} See @Tesserex & @erkangur
I think this is a good explanation of permutations and combinations:
Easy Permutations and Combinations Better explained.
It goes step by step until you discover how to make the calculations.