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  • A 6-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 2 and have exactly 3 even digits?

A 6-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 2 and have exactly 3 even digits?

Option: 1

25,000

 


Option: 2

60,000

 


Option: 3

10,000

 


Option: 4

40,000


Answers (1)

best_answer

To calculate the number of different 6-digit numbers that can be formed using the digits 0-9, where repetition is allowed, and the number must be divisible by 2 with exactly 3 even digits, we can consider the following:

Since the number must be divisible by 2 , the last digit must be even. We have 5 even digits (0,2,4,6, 8) to choose from for the last digit.

For the remaining 5 digits, we need to select 2 even digits and 3 odd digits. We have 5 even digits and 5 odd digits to choose

from, so we can select the 2 even digits in \mathrm{C(5,2)} ways and the 3 odd digits in \mathrm{C(5,3)} ways.

Once we have selected the digits, we can arrange them in the remaining 5 positions in 5 ! ways.

Therefore, the total number of different 6-digit numbers that can be formed with the given conditions is:

C(5,2) \times C(5,3) \times 5 ! \times 5=10 \times 10 \times 120 \times 5=60,000 .

Therefore, there are 60,000 different 6-digit numbers that can be formed, satisfying the conditions of being divisible by 2 and having exactly 3 even digits.

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