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  • A 5-digit number is to be formed using the digits 0-9, where repetition is allowed. How many different numbers can be formed if the number must be divisible by 3 and have

A 5-digit number is to be formed using the digits 0-9, where repetition is allowed. How many different numbers can be formed if the number must be divisible by 3 and have exactly 2 odd digits?

 

Option: 1

2500


Option: 2

3400


Option: 3

3100


Option: 4

7800


Answers (1)

best_answer

To calculate the number of different 5 -digit numbers that can be formed using the digits $0-9$, where repetition is allowed, the number must be divisible by 3 , and exactly 2 odd digits are used, we can consider the following:

Since the number must be divisible by 3 , the sum of its digits must be divisible by 3 .

Case 1: The sum of the digits is divisible by 3 and the number has 2 odd digits.

In this case, we need to determine the possible arrangements of the digits. There are 5 odd digits (1, 3,5,7,9) and 5 even digits (0,2,4,6,8) to choose from.

The number of ways to choose 2 odd digits from the 5 odd digits is \mathrm{C(5,2)=10.}

Once we have chosen the 2 odd digits, we can arrange them in the number in 2 ! ways.

The remaining 3 digits can be chosen from the 5 even digits in \mathrm{5^3} ways (since repetition is allowed).

Therefore, the total number of different numbers that can be formed in this case is:

10 \times 2 ! \times 5^3=10 \times 2 \times 125=2500.

Case 2: The sum of the digits is divisible by 3 , but the number has 4 odd digits.

In this case, we need to determine the possible arrangements of the digits. There are 5 odd digits and 5 even digits to choose from.

The number of ways to choose 4 odd digits from the 5 odd digits is \mathrm{C(5,4)=5.}

Once we have chosen the 4 odd digits, we can arrange them in the number in 4 ! ways.

The remaining digit can be chosen from the 5 even digits in 5 ways.

Therefore, the total number of different numbers that can be formed in this case is:

5 \times 4 ! \times 5=600 .

Therefore, the total number of different 5-digit numbers that can be formed, satisfying the conditions of being divisible by 3 and having exactly 2 odd digits, is:

\mathrm{ 2500+600=3100 . }

Hence, there are 3100 different 5-digit numbers that can be formed with the given conditions.

Posted by

SANGALDEEP SINGH

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