## 4064465

You are given the following seven-digit number: $4064465$. How many different seven-digit numbers can be made by rearranging the digits of $4064465$?
Source: mathcontest.olemiss.edu 4/30/2012

SOLUTION
Consider the set of seven digits $\left \{0,4,4,4,5,6,6\right \}$. We have seven places to put the seven digits. We can put the digits $\left \{4,4,4,5,6,6\right \}$ in any place but we cannot put digit $\left \{0\right \}$ in the leading place because we want to make seven-digit numbers.

Though we can begin with any digit, let’s start with digit $\left \{0\right \}$ which occupies one place.
How many ways can we choose one place out of six (not seven because no leading zero) to put the digit $\left \{0\right \}$?
$\binom{6}{1}$

There are six places remaining to put the remaining six digits $\left \{4,4,4,5,6,6\right \}$.
How many ways can we choose three places out of six to put the digits $\left \{4,4,4\right \}$?
$\binom{6}{3}$

There are three places remaining to put the remaining three digits $\left \{5,6,6\right \}$.
How many ways can we choose one place out of three to put the digit $\left \{5\right \}$?
$\binom{3}{1}$

There are two places remaining to put the remaining two digits $\left \{6,6\right \}$.
How many ways can we choose two places out of two to put the digits $\left \{6,6\right \}$?
$\binom{2}{2}$

Total ways = $\binom{6}{1}\times\binom{6}{3}\times\binom{3}{1}\times\binom{2}{2}$
$=6\times 20\times 3\times 1$
$=360$

Answer: $360$