A keycode must contain 4 letters and 3 numbers. The letters may be any letter of the alphabet. The numbers should be any number from 0 to 5. How many different keycode combinations are there?
I have the above problem and the more I look at it the more confused i get!
If no repetitions are allowed then I am pretty sure the the answer is
26C4 * 6C3 * 7!
But let's assume repetition is allowed.
I'll choose the letters first there are $26^4$ ways.
If the numbers are all the same then there are 6 ways of choosing them.
So the number of selections might be $26^4*6*7!/3!$
If the first 2 numbers chosen are the same there are 6*1*5=30 ways to choose them
If the last 2 numbers are the same there are 6*5*1=30 ways to choose them
If the 1st and 3rd are the same then there are 6*5*1=30 ways to choose them
SO there are 90 ways that 2 numbers can be chosen the same and the 3rd one different.
So the number of selections might be $ 26^4*90*7!/2!$
If the numbers are all different then the number of possibilities is 6*5*4=120
So the number of selections might be $26^4*120*7!$
Actually I think I might need to go through the same rigmarole for the letters. Yes I think that i do.
Letters all different = 26*25*24*23=358800
letters 2 the same= 26*1*25*24*6= 93600
letters 3 the same=26*1*1*25*4=2600
Letters all the same=26
Add all these and I get 455026
Now I know it should add up to $26^4=456976$ so I am short a few possibilities here. WHY AM I SHORT???
See i do not have a clue what I am doing!
Can anyone help set me straight - there are a number of different aspects to this question.
Actually 3 of us have been struggling with this question. At one stage all of us had the same answer but I am almost positive it was wrong. You call us the 3 stooges I suppose.
Any ideas will be greatly appreciated.