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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.

nP_{r} = n(n − 1)(n − 2) ⋯ (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

nΠ_{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, …, and r are the same, the number of permutations in which they are arranged in one column is

\binom{n}{p, q, …, r} = \frac{n!}{p!q! ⋯ r!}, p + q + ⋯ + 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 _{6}P_{mboxItcanbeexpressedas3}. 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 ·_{5}P_{2} = 5 · 5 · 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