Is this formula for calculating the ways to arrange the 5 digits correct? If it is, why?

Question

I am trying to solve how many ways are these to arrange 6,7,8,9,9, these 5 digits. I asked my friend, he told me that I can use $\frac{5!}{2!}$ to do this. I have no idea how this work, could you guys please explain to me? I knew that we could have 5! way to manipulating these digits, my problem is how do we know there would be 2! repeated.

Thank you so much for your reply

Answer 1

Suppose we had $6,7,8,9,\color{red} 9$

Now it is easier, because we can differentiate the 9's

There are $5!$ ways to arrange $5$ objects.

And for each of these arrangements we can swap the 9's and a colorblind person would know the difference.

$\frac {5!}{2}$

Answer 2

Apply the rule of product

• Pick where in the resulting permutation the $6$ goes (five options)

• Pick where in the resulting permutation the $7$ goes (four remaining options)

• Pick where in the resulting permutation the $8$ goes (three remaining options)

• Let the final positions remaining be occupied by both $9$'s

This gives a total of $5\times 4\times 3$ (which is clearly equal to $\frac{5!}{2!}$)

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Even more generally, you might choose to use multinomial coefficients for larger problems.

Answer 3

In total, there are 5! ways of ordering 5 objects in a row. Since two of the digits are the same $9,9$, each number we get appears exactly 2! times (second time when the first 9 and the second 9 flip places).

That's why you get $5!/2!$.