Animated Solution for Mathematics - Permutations and Combinations: Total number of six-digit numbers in which only and all the five digits 1, 3, 5, 7 and 9 appear, is

Imagine you are standing before a digital vault. You have a keypad with only five buttons: $1,3,5,7,$ and $9$. Your task is to create a six-digit code using only and all of these five digits.

We have $6$ slots to fill, but only $5$ distinct digits to work with. Because we are forbidden from using any digits outside our set of $\{1,3,5,7,9\}$, we are forced to reach back into our pool and pick one of the digits we have already used.

This is the Pigeonhole Principle in its most elegant form: if you have $6$ slots and $5$ digits, at least one digit must repeat. Exactly one digit will appear twice, while the other four appear exactly once.

Before we can arrange these digits, we must decide which digit receives the privilege of being the 'repeater'. We have $5$ candidates: $1,3,5,7,$ and $9$.

We need to choose exactly one of them to appear twice. This is a simple selection problem: how many ways can we choose $1$ digit out of $5$? The answer is ${}^5{C}_1$, which is simply $5$.

For instance, if we choose the digit $3$, our pool of digits for the six-digit number becomes $\{1,3,3,5,7,9\}$.

Now that we have our set of $6$ digits, we need to arrange them into our $6$ slots. If all $6$ digits were distinct, the answer would be $6!$.

However, we have a complication: two of our digits are identical. To account for this, we use the formula for permutations of a multiset:

Here, $n=6$ and $p=2$. Thus, the number of unique arrangements for a fixed set is:

We have two independent stages. First, we select the repeating digit ($5$ ways). Second, we arrange the resulting set of digits ($360$ ways).

By the Fundamental Principle of Counting, we multiply these two stages together to find the total number of possible six-digit numbers:

The total number of possible six-digit codes is 1800.