Question

How many different license plate numbers can be made using 2 letters followed by 4 digits selected from the digits 0 through \(9,\) if: (a) Letters and digits may be repeated? (b) Letters may be repeated, but digits may not be repeated? (c) Neither letters nor digits may be repeated?

Short answer

(a) \(26^2 \times 10^4\), (b) \(26^2 \times (10 \times 9 \times 8 \times 7)\), (c) \((26 \times 25) \times (10 \times 9 \times 8 \times 7)\)

Step-by-step solution

Calculate for case (a)

In this case, both letters and digits can be repeated. Considering that there are 26 letters in the English alphabet and each license plate has 2 letters, the total number of combinations for the letters is calculated by: \[ 26 \times 26 = 26^2 \] For the digits, each license plate has 4 digits, so the total number of combinations is: \[ 10 \times 10 \times 10 \times 10 = 10^4 \] Finally, multiply the combinations of letters and digits together: \[ 26^2 \times 10^4 \]

Calculate for case (b)

In this case, letters may be repeated, but digits may not be repeated. The total number of combinations for the letters is still: \[ 26^2 \] For the digits, we must select 4 unique digits from the 10 available digits. The number of permutations of 4 digits from a set of 10 is calculated by: \[ 10 \times 9 \times 8 \times 7 \] Multiply the combinations of letters and digits together: \[ 26^2 \times (10 \times 9 \times 8 \times 7) \]

Calculate for case (c)

In this case, neither letters nor digits may be repeated. The number of permutations for 2 letters from 26 available letters is calculated by: \[ 26 \times 25 \] The number of permutations for 4 digits from 10 available digits is still: \[ 10 \times 9 \times 8 \times 7 \] Multiply the combinations of letters and digits together: \[ (26 \times 25) \times (10 \times 9 \times 8 \times 7) \]

Key concepts

permutations

Permutations are arrangements of objects where the order is important. When we talk about permutations, we're looking at how many ways we can arrange a set of objects.
For example, if you have three letters: A, B, and C, and you want to know how many ways you can arrange all three letters, you consider each different order.
This means ABC, ACB, BAC, BCA, CAB, and CBA are all unique permutations, giving us a total of 6 permutations, which can be computed as: \[3! = 3 \times 2 \times 1 = 6\]
In our exercise, when we deal with license plates where neither letters nor digits can be repeated, we use permutations. For instance, choosing 2 letters out of 26 available letters, the permutations formula is applied: \[26 \times 25\] since the order in which we select the letters matters, and once a letter is used, it can't be reused.
Similarly, if we choose 4 digits out of 10 without repetition, the permutations method leads us to: \[10 \times 9 \times 8 \times 7\].

combinations

Combinations are selections of objects where the order does not matter. Unlike permutations, the arrangement sequence isn’t important in combinations.
If you have the same set of three letters: A, B, and C, and you want to know how many ways you can select 2 letters, combinations focus only on the groups, not the order.
This means AB, BA, AC, CA, BC, and CB are counted as AB, AC, and BC once, giving us just 3 combinations. This is computed by the combinations formula: \(\binom{3}{2} = \frac{3!}{2!(3-2)!} = 3\).
Though in our exercise, we primarily deal with permutations, understanding combinations helps us in contexts where order doesn't play a role.
It becomes handy in different scenarios where selection without regard to order is the goal, such as choosing team members from a pool.

license plate calculation

To calculate possible license plate numbers, we blend principles from permutations and combinations.
For each specific scenario in our exercise, we need to discern if repetition is allowed and whether order matters.
In the exercise:

• Case (a): Both letters and digits may be repeated. This is the simplest scenario where all characters can repeat:
  \(26^2 \times 10^4\)
• Case (b): Letters may be repeated, but digits cannot be repeated. Letters are calculated the same way as in case (a), but digits follow the permutation rule without repetition: \(26^2 \times (10 \times 9 \times 8 \times 7)\)
• Case (c): Neither letters nor digits may be repeated. Both letters and digits follow the permutation rule without repetition:
  \((26 \times 25) \times (10 \times 9 \times 8 \times 7)\)

Understanding which principles apply in each scenario helps students choose the right formulas to accurately calculate working combinations.

repetition in counting

Repetition plays a crucial role in determining the number of permutations or combinations.
When repetition is allowed, we have more options because each selection is independent of the previous ones.
In our exercise, case (a) where repetition of both letters and digits is allowed, results in a straightforward multiplication problem: \[26 \times 26 \times 10 \times 10 \times 10 \times 10 = 26^2 \times 10^4\]On the other hand, without repetition, options decrease after each selection.
For example, in case (b), after choosing the first digit, the choice for the next digit cannot include the already selected digits:

• 1st digit: 10 options
• 2nd digit: 9 options, and so forth: \(10 \times 9 \times 8 \times 7\).

So, remembering whether repetition is permitted helps define if we are looking at simple multiplication or decrementing options for each subsequent choice.