Question

How many license plates can you build with 3 letters and 3 numbers with repetitions? You also need the additive principle.

Answer 1

26*26*26*10*10*10
26 possible letters, 10 possible numbers.
You can start with 26 different letters. For each, the next can be 26 different letters. For each of those, 26 possible letters.
Then, for each of those, 10 possible numbers and so on

Answer 2

26*26*26*10*10*10
26 possible letters, 10 possible numbers.
You can start with 26 different letters. For each, the next can be 26 different letters. For each of those, 26 possible letters.
Then, for each of those, 10 possible numbers and so on

Answer 3

that cant be right because there are only 3 letters and 3 numbers

Answer 4

then you need to calculate possible ordering and remove new orders that duplicate. Try it and post if you get stuck

Answer 5

where is the 26 and 10 coming from

Answer 6

26 letters in the alphabet, 10 different digits

Answer 7

you can only use 3 letters


and 3 numbers

Answer 8

it doesnt say 26 numbers

Answer 9

26 letters

Answer 10

I think that means 3 of the characters on the plates are letters, and there are 26 different choices for each letter slot. Maybe I'm misunderstanding, but that is what the question would normally ask

Answer 11

How many license plates can you build with 3 letters and three numbers with repetiion

Answer 12

so im thinkink its sayying like letters {a,b,c} and numbers {1,2,3}

Answer 13

so MKV123 is 3 of each, but so is JHC567. Both are 3 of each, as is 1T2U4B

Answer 14

no

Answer 15

the 26*26*26 thing was right

Answer 16

but i have to use the additive princible

Answer 17

yeah, because you can start with a letter or a number which increases the permutations, then reduce that for the reorderings that end up duplicating a permutation

Answer 18

can you show me how to do it im not understanding what to do

Answer 19

so in your book, you have stuff like 3P26, 10C12 and that sort of thing.
the 26 P 3 rather... 26 things taken 3 at a time. There will be a formula with it

Answer 20

but we can have repeats, so we are really using 26 P 1 three times

Answer 21

which gives us the 26*26*26

Answer 22

yeah i figured that out im not sure where to go from there

Answer 23

so you do 26*26*26*10*10*10 first. Then we can start with any of the three letters or numbers, yes? So 6 choices, then 5 for the next slot, 4 choices for the third slot and so on, and that ends up being 6! which we multiply our already big number by

Answer 24

We would be done here except that we don't want to double count duplicates.
if I have WWC567 and I swap the two Ws, then I don't actually have a new license number

Answer 25

So we need to figure out how many times that happens

Answer 26

For any given arrangement, only swapping around letters or numbers with the same type could duplicate us. If I swap a letter and a number, I'm going to have a new arrangement.

Answer 27

Is there anyone else that can jump in here? I need to run. I didn't think this would take a while.

Answer 28

This bit is where you use the combinations formulas to eliminate repeats

Answer 29

I'm sorry, I have to meet friends in 10 minutes. Reread the examples in your book and see how far you get. Hopefully someone else will jump in!