Am I using the right equations? Determine the number of possible outcomes when making a 7-character license plate using the letters of the alphabet and the digits 1-4 if the first 3 characters must be non-repeating letters and the remaining 4 are digits that may repeat. 26C3 x 16C4
so there are 26 + 4 choices... the 1st 3 without replacement 30 x 39 x28 x 30C4
I would think 26*25*24*4^4
that correct... 26 x 25x 24 x4^4
I'm pretty confused right now...
Rationale for the above answer: The first letter can be any of 26. The next can be any of the remaining 25, the next can be any of the remaining 24. Each of the four digits can be any of the four digits available, so those contribute a factor of 4^4.
I get 3993600.
That's the right answer (have the answer key) I'm just trying to figure out how you got there... I get the 26*25*24 etc, which is why I put 26C3... or is there no etc, and it's just 26 x 25 x 24?
\[26\cdot25\cdot24=\,_{26}P_{3}\]
-sigh- I'm so confused. I think I'll just leave this question for tutoring ^^;
\[26\cdot25\cdot24\ne\,_{26}C_{3}=\frac{26\cdot25\cdot24}{3\cdot2\cdot1}\]
True, but why is that example a permutation?
the order that you pick the letter matters
... true :o But then the numbers are C, right? (I can't remember the name of it atm, too tired, but I know what it is lol)
for 'C' the order does not matter
And the numbers may repeat, so the order doesn't matter, right?
the order does matter.... you get 4x4x4x4
But why...?
why the order matters or why it is 4x4x4x4?
Both ^^;
are ABC1234 and ABC4321 the same license plate?
No...
so the order of the numbers matters
x.x... so why 4^4?
and since you can have repeated numbers..you have 4 choices for the first number , 4 for the second, 4 for the third, and 4 for the fourth. Thus 4x4x4x4 total
Ah.... that's true
4^4 = 4096, right?
\[4^4=256\]
x_x I don't have my graphing calc, sorry. So do I do 256 x 26P3?
put 256*26*25*24 into google
Mhmm... that's the answer
26*25*24 = 26P3, right...?
yes
Okay :) Thank you so much!
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