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How many three-digit multiples of 5 have three different digits and at least one prime digit?

Sagot :

Using the Fundamental Counting Theorem, it is found that there are 124 three-digit multiples of 5 that have three different digits and at least one prime digit.

What is the Fundamental Counting Theorem?

It is a theorem that states that if there are n things, each with [tex]n_1, n_2, \cdots, n_n[/tex] ways to be done, each thing independent of the other, the number of ways they can be done is:

[tex]N = n_1 \times n_2 \times \cdots \times n_n[/tex]

Multiples of 5 finish at 0 or 5, hence the parameters to find the number of three-digit multiples of 5, with different digits are:

[tex]n_1 = 9, n_2 = 8, n_3 = 2[/tex]

And the number is:

N = 9 x 8 x 2 = 144.

With no prime digits, 2, 3, 5 and 7 cannot be used, hence the parameters are:

[tex]n_1 = 5, n_2 = 4, n_3 = 1[/tex]

Hence 20 of the numbers have no prime digits, and 144 - 20 = 124 have at least one prime digit.

More can be learned about the Fundamental Counting Theorem at https://brainly.com/question/24314866

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