Calculating Permutations where n < r

Question

I have a real life problem that sounds like word problem from an Algebra test, but I've haven't been able to find an example of how to solve it.

My sister forgot the security code to the garage. The security code is 4 digits (order matters). She remembers that the code consists of a 6, 5, and 2 and that one digit repeats somewhere in the sequence. I want to figure out the total number of possible security codes

Every example of permutations calculations I have been able to find require that the total number of objects being selected from is greater than the number of selections (i.e. a 4 digit code made up of the digits 0-9). Also, the examples that allow for repetition seem to allow all digits to repeat not just one.

Answer 1

Apply multiplication principle

• Pick which of the digits (6, 5, or 2) is the digit that is repeated (3 options)

• Pick the location of the smallest non-repeated digit (4 options)

• Pick the location of the largest non-repeated digit (3 options)

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Remember that the multiplication principle of counting can be reworded to say

If you can come up with a sequence of choices whose choices uniquely describe every outcome/arrangement which you want to count in exactly one way such that the number of options available at each choice does not change based on choices made in previous steps then the total number of outcomes is the product of the number of choices available at each step.

Answer 2

You have three choice of the digit to duplicate. For each of those you have $\frac {4!}{2!}$ orders, the $4!$ from putting four things in order and the $2!$ because we can reorder two of them and not care. $3 \cdot \frac {4!}{2!}=36$

Answer 3

There are $3$ choices for the repeating digit.

There are $\binom{4}{2}$ positions (from $1$ to $4$) for the repeating digit.

Once the repeating digit is placed, there are two ways to place the remaining two digits.

Hence the number of possible arangements is $$(3)\binom{4}{2}(2) = (3)(6)(2)=36$$