How many different five-digit numbers divisible by five can be written with the numbers 0,2,3,5,7?


Answer from: Vyrvidub Svyatogryzych:
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Let's use the basic rule of combinatorics:
Let there be k groups of elements, and the i-th group consists of ni-elements. Let's choose one element from each group. Then the general number N of ways, by which it is possible to make such a choice, is determined by the ratio N=n1*n2*n3*...*nk.

n1 = 4 (You can choose any number except 0, it is not allowed to start with)
n2 = 5 (Any number will do)
n3 = 5
n4 = 5
n5 = 2 (You can only choose 0 and 5, otherwise it is not a multiple of 5)
N = 4 * 5 * 5 * 5 * 2 = 4 * 25 * 10 = 100 * 10 = 1000


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