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Korvikt [17]
1 year ago
8

How many 3 digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9 which are divisible by 5 and none of the digits is re

peated?.
Mathematics
1 answer:
Neko [114]1 year ago
5 0

The set of numbers given can form 20 3-digit numbers that are divisible by 5.

Permutation is the number of ways an <em>r </em>digit number can be arranged from a set of<em> n </em>numbers. Also, this type of arrangement gives importance to the order or sequence of numbers.

Given: r = 3 digit number, n = 6 numbers in the set (2, 3, 5, 6, 7, and 9)

Since none of the digits should be repeated, then the number of choices each time decreases. In addition, the number should be divisible by 5, which means that the only number that should be at the end is 5. This leaves only five choices for the first number and four choices for the second number. We can now determine how many 3-digit numbers that are divisible by 5 can be formed from the set of numbers given.

Number of 3-digit numbers divisible by 5 = (first number)*(second number)*(third number)

Number of 3-digit numbers divisible by 5 = 5*4*1 = 20

To learn more about permutation, please refer to the link brainly.com/question/4658834.

#SPJ4

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