ePermutationComb

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 from different pieces and arrange them in one column.

$\displaystyle {}_{n}P_{r} = n(n-1)(n-2)\cdots (n-r+1) = \frac{n!}{(n-r)!}$
Especially, ${}_{n}P_{n} = n!$ ，．
permutation with repetition The number of permutations that allow repetition from different pieces is

$\displaystyle {}_{n}\Pi_{r} = n^{r}$
Circular permutation The number of permutations that arrange different things in a circle is

$\displaystyle (n-1)!$
Permutations containing the same Of the items, when , , $\ldots$ , and are the same, the number of permutations in which they are arranged in one column is

$\displaystyle {n \choose {p,q,\ldots,r}} = \frac{n!}{p! q! \cdots r!}, \ p+q+\cdots +r = n$

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

$\displaystyle 120 + 100 = 220 \$ ways $\displaystyle$

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