The total number of numbers, lying between 100 and 1000 that can be formed with the digits 1, 2, 3, 4, 5, if the repetition of digits is not allowed and numbers are divisible by either 3 or 5, is _____________.
Answer (integer)
32
Solution
The numbers are lying between 100 and 1000
then each number is of three digits.
<br><br>The possible combination of 3 digits numbers
are
<br><br>1, 2, 3; 1, 2, 4; 1, 2, 5; 1, 3, 4; 1, 3, 5; 1, 4, 5;
2, 3, 4; 2, 3, 5; 2, 4, 5; and 3, 4, 5.
<br><br>The possible combination of numbers which are divisible by 3 are 1, 2,
3; 3, 4, 5; 1, 3, 5 and 2, 3, 4.
<br>(If sum of digits of a number is divisible by 3 then the number is divisible by 3)
<br><br>$\therefore$ Total number of numbers = 4 × 3! = 24
<br><br>The possible combination of numbers divisible by 5 are 1, 2, 5; 2, 3, 5; 3, 4, 5; 1, 3, 5;
1, 4, 5 and 2, 4, 5.
<br>(If the last digit of a number is 0 or 5 then the number is divisible by 5)
<br><br>$\therefore$ Total number of numbers = 6 × 2! = 12
<br><br>The possible combination of number divisible by both 3 and 5 are 1, 3, 5 and 3, 4, 5.
<br><br>$\therefore$ Total number of numbers = 2 $\times$ 2! = 4
<br><br>$\therefore$ Total required number = 24 + 12 - 4 = 32
About this question
Subject: Mathematics · Chapter: Permutations and Combinations · Topic: Fundamental Counting Principle
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