Discrete Mathematics - Permutations/Combinations?

In summary, the conversation discussed the task of issuing automobile license plates in a certain state, with the requirement of 2 letters followed by 3 digits on each plate. The question was how many different license plates could be issued if letters and digits cannot be repeated. After discussing the rules of factorials, it was determined that the correct solution is to multiply the number of choices for each character, resulting in a total of 468,000 possible license plates.
  • #1
aliaze1
174
1

Homework Statement



A certain state issues a series of automobile license plates such that each license plate
must have 2 letters followed by three digits. An example license plate would be AD 025 .
If the letters and the digits cannot be repeated, how many different license plates can be
issued by the state?

(a) 468,000 (b) 486,720 (c) 46,800 (d) 1,300 (e) 67,600

Homework Equations



Rules of factorials?
(ex: 5! = 5*4*3*2*1)

The Attempt at a Solution



How would I go about this?

I was thinking (26*25) + (10*9*8), but that is not an available option

Thanks
 
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  • #2
well think of it like this: If the first 2 must be letters...how many choices do you have to pick the first letter?26 right...and if you pick one letter out then you have 25 remaining..so from the 25 you can pick one for the 2nd letter. and then for the rest of numbers it should be picking 3 numbers from 10..where the order is important
 
  • #3
Why did you choose to add the probabilities? Try multiplying.
 
  • #4
Integral said:
Why did you choose to add the probabilities? Try multiplying.

I originally thought they were separate things...

Thanks!
 
  • #5
I got 468,000 by multiplying 26*25*10*9*8.
 
  • #6
me2

bondgirl007 said:
I got 468,000 by multiplying 26*25*10*9*8.

thanks!
 

Q1. What is the difference between a permutation and combination in discrete mathematics?

A permutation is an arrangement of objects where order matters, while a combination is a selection of objects where order does not matter. In other words, in a permutation, the order of the elements matters, while in a combination, the order does not affect the result.

Q2. How do you calculate the number of permutations of a set?

The number of permutations of a set, denoted by nPr, can be calculated using the formula n!/(n-r)!, where n is the total number of objects in the set and r is the number of objects being selected. For example, if we have a set of 5 objects and we want to find the number of permutations of 3 objects, we would use the formula 5!/(5-3)! = 5!/2! = 60.

Q3. How do you calculate the number of combinations of a set?

The number of combinations of a set, denoted by nCr, can be calculated using the formula n!/r!(n-r)!, where n is the total number of objects in the set and r is the number of objects being selected. For example, if we have a set of 5 objects and we want to find the number of combinations of 3 objects, we would use the formula 5!/(3!(5-3)!) = 5!/3!2! = 10.

Q4. What is the difference between a permutation with repetition and a combination with repetition?

A permutation with repetition is an arrangement of objects where some elements may be repeated, while a combination with repetition is a selection of objects where some elements may be repeated. In other words, in a permutation with repetition, the same element can appear multiple times in the arrangement, while in a combination with repetition, the same element can be selected multiple times.

Q5. How do you solve problems involving permutations and combinations?

To solve problems involving permutations and combinations, it is important to first identify whether the problem involves ordering or selection, and whether repetition is allowed. Then, you can use the appropriate formula to calculate the number of permutations or combinations. Additionally, it is important to carefully read and understand the problem to determine whether any additional steps or adjustments need to be made in the calculation.

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