anonymous
  • anonymous
How many license plates can you build with 3 letters and 3 numbers with repetitions? You also need the additive principle.
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
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
anonymous
  • anonymous
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
anonymous
  • anonymous
that cant be right because there are only 3 letters and 3 numbers

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anonymous
  • anonymous
then you need to calculate possible ordering and remove new orders that duplicate. Try it and post if you get stuck
anonymous
  • anonymous
where is the 26 and 10 coming from
anonymous
  • anonymous
26 letters in the alphabet, 10 different digits
anonymous
  • anonymous
you can only use 3 letters and 3 numbers
anonymous
  • anonymous
it doesnt say 26 numbers
anonymous
  • anonymous
26 letters
anonymous
  • anonymous
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
anonymous
  • anonymous
How many license plates can you build with 3 letters and three numbers with repetiion
anonymous
  • anonymous
so im thinkink its sayying like letters {a,b,c} and numbers {1,2,3}
anonymous
  • anonymous
so MKV123 is 3 of each, but so is JHC567. Both are 3 of each, as is 1T2U4B
anonymous
  • anonymous
no
anonymous
  • anonymous
the 26*26*26 thing was right
anonymous
  • anonymous
but i have to use the additive princible
anonymous
  • anonymous
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
anonymous
  • anonymous
can you show me how to do it im not understanding what to do
anonymous
  • anonymous
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
anonymous
  • anonymous
but we can have repeats, so we are really using 26 P 1 three times
anonymous
  • anonymous
which gives us the 26*26*26
anonymous
  • anonymous
yeah i figured that out im not sure where to go from there
anonymous
  • anonymous
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
anonymous
  • anonymous
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
anonymous
  • anonymous
So we need to figure out how many times that happens
anonymous
  • anonymous
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.
anonymous
  • anonymous
Is there anyone else that can jump in here? I need to run. I didn't think this would take a while.
anonymous
  • anonymous
This bit is where you use the combinations formulas to eliminate repeats
anonymous
  • anonymous
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!

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