How many different 5-digit numbers can be formed from digits 2, 2, 7, 7, 9?

Question

As I am solving some basic combinatorial problems today, I found out this problem:

How many different 5-digit numbers can be formed from digits 2, 2, 7, 7, 9?

Can someone guide me to a solution for this problem? It's probably something related to multisets. I'm beginner in this area. Thanks!

Answer 1

So, you have five spots to put $2,2,7,7,9$ into. First, you pick two spots out of the five for the $2$'s, for which there are $\binom{5}{2}$ ways. Now, there are three spots remaining, and I want to pick two spots out of the three for the $7$'s. For this there are $\binom{3}{2}$ ways. (Note that if you were to pick one spot for the $9$, $\binom{3}{1} = \binom{3}{2}$. Hence, there are $$\binom{5}{2}\binom{3}{2} = \frac{5!}{2!2!1!} = 30$$ ways of doing this.

Answer 2

$ 3^5 $ with repetition ( $\{2,7,9\}\times\{2,7,9\}\times\{2,7,9\}\times\{2,7,9\}\times\{2,7,9\}$, $99777$ for example )

$5!/(2!\cdot2!) = 30$ without repetition (except $2$ or $7$, $27297$, for example)

Answer 3

It is just asking you indirectly that

How many arrangements can be made with the given numbers.

If the numbers would have been different than it would have simply been 5! without any repetition, But since 2 and 7 are repeated twice it can be arranged in 5!/2!2!.