Hello all , can you please help me solve this question :

How many license plates of 3 symbols (letters and digits) can be made using at least one letter in each? ( repetition is allowed)

 

So you are drawing sets of 3 from a pool of 36 items with replacements?

Think about choosing the 3rd element once the first two have been chosen. How many options are there?

 

I solve it like this (26.36.36) + (36.26.36) + (36.36.26)

Is it correct like this?

 

Ahh, I missed that each combination must include a letter. Your calculation looks reasonable. It should produce the same result if you subtract 10•10•10 from 36^3?

 

Ahh, I missed that each combination must include a letter. Your calculation looks reasonable. It should produce the same result if you subtract 10•10•10 from 36^3?

I solve it like this in first but after , I think this solution is correct :
(26.36.36) + (36.26.36) + (36.36.26) = 101.088 different plates .
or what ?

 

36^3 = 46,656
10^3 = 1,000

So I don't think your approach is correct now that I have checked it.

There are formulas for this, you know. One thing you want to learn is where to find them and how to use them. You might even be expected to remember a few of them and produce them on an exam.

 

36^3 = 46,566
10^3 = 1,000

So I don't think your approach is correct now that I have checked it.

There are formulas for this, you know. One thing you want to learn is where to find them and how to use them. You might even be expected to remember a few of them and produce them on an exam.

Thanks a lot for help , so are you sure about your answer ?

 

I have no dog in the fight. You're the one that needs to be sure.

This might help.
https://www.mathsisfun.com/combinatorics/combinations-permutations-calculator.html

 

I have no dog in the fight. You're the one that needs to be sure.

This might help.
https://www.mathsisfun.com/combinatorics/combinations-permutations-calculator.html

But why my solution is wrong ?

 

It incorrectly double and triple counts some of the permutations. In your summing, many permutations can occur in each of the three components of the sum.

 

It incorrectly double and triple counts some of the permutations. In your summing, many permutations can occur in each of the three components of the sum.

Ok , thank you so much really appreciate it .

 

Ok , thank you so much really appreciate it .

It's not unusual to be confused by this stuff! If you want to blow your mind, look up the "Monty Hall problem".

 

But why my solution is wrong ?

I think Waynes approach is correct.

You are double or triple counting sequences.
For example your first term includes KJ3.
Your second term includes KJ3 as well.
KKK could be included in all three terms - and it would be a prohibited sequence in most states.

 

Now I get it , thanks so much for help .