Permutations ?

  • Thread starter ZeroPivot
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  • #1
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Homework Statement



Lets say a car plate has 6 positions, three of them are numbers and 3 of them are alphabetic.
e.g. ABC 123. I need to figure out how many possible permutations can occur using all the numbers and alphabets BUT i cannot use any number or alfabet twice in an permutation eg. AAB 123 or ABC 112 is forbidden.


Homework Equations





The Attempt at a Solution



first i split the problem, first the numbers then the alfabet:

NUMBERS:
so N1 is the number of numbers i can use which is 10 obviously. 0-9
K is the number of positions which is 3.

N1!/(N1-K)=10!/(10-3)!

ALFABET:
so N2 is the number of alfabets which is 26, K is the number of positions 3.

N2!/(N2-K)!=26!/(26-3)!

The ANSWER:

(N1!/(N1-K))*(N2!/(N2-K)!) = (10!/(10-3)!)*(26!/(26-3)!)

is the answer correct?
 

Answers and Replies

  • #2
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That is correct, assuming every car plate has to use 3 letters and 3 digits. You can calculate that number as a final result.
 
  • #3
54
0
That is correct, assuming every car plate has to use 3 letters and 3 digits. You can calculate that number as a final result.

thanks, does the answer exclude the permutations where the same number or digit happens in all three slots? e.g. AAA 123 or ABC 111 ?
 
  • #4
CAF123
Gold Member
2,950
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thanks, does the answer exclude the permutations where the same number or digit happens in all three slots? e.g. AAA 123 or ABC 111 ?
Yes, what you have is $$3! \cdot{10 \choose 3} \times 3! \cdot {26 \choose 3}$$ which means, in the first term you select 3 distinct objects from the group of size 10. Each choice of three objects then has 6 possible rearrangements. Similar for the other term.

If you are allowed to use the same objects more than once, then clearly the number of possible permutations will increase.
 

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