Example Find out how many 4-digit integers are multiples of 5 when extracting and arranging 7 numbers from 0 to 6．

Answer When considering how many ways a certain thing can happen, the number of ways it happens is called number of possible outcomes.

• permutation The number of permutations that take $r$ from different $n$ pieces and arrange them in one column.
  $${}_{n}P_{r} = n(n-1)(n-2)\cdots (n-r+1) = \frac{n!}{(n-r)!}$$
  Especially, ${}_{n}P_{n} = n!$，$0! = 1$．
• permutation with repetition The number of permutations that allow repetition $r$ from different $n$ pieces is
  $${}_{n}\Pi_{r} = n^{r}$$

• Circular permutation The number of permutations that arrange different $n$ things in a circle is
  $$(n-1)!$$

• Permutations containing the same Of the $n$ items, when $p$, $q$, $\ldots$, and $r$ are the same, the number of permutations in which they are arranged in one column is
  $${n \choose {p,q,\ldots,r}} = \frac{n!}{p! q! \cdots r!}, \ p+q+\cdots +r = n$$

Answer to the problem

First, note that multiples of 5 always have a first digit of 0 or 5. Next, consider the case where the first digit is 0 and the case where it is 5.

When the first digit is 0:

Thousands, hundreds, and tens can use numbers from 1 to 6 once, so 3 out of 6 are permuted, so ${}_6P_{\\ mbox{It can be expressed as 3}}$. Therefore 120 ways

When the first digit is 5:

Since 0 cannot be used for the thousand's place, the thousand's place can be used in 5 ways other than 0 and 5, and the hundred's and tens”s can be used once for any number other than the numbers used for the one's and thousands's place. Therefore, $5\cdot {}_5P_{2} = 5\cdot 5 \cdot 4 = 100$ ways.

Here, when the ones digit is 0 and 5, it does not occur at the same time, so in total

$$120 + 100 = 220 \$$   ways$$$$