Number of choices for the given set of digits {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} to make a pin of four digit with the condition of last digit to be even and second to last must be odd is equal to 2,500.
As given in the question,
Given set of digits is :
{0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
Number of digits required to form a pin = 4 digits
Condition given:
Last digit should be even :
Last place can be filled by 0,2,4,6,8
5 choices to fill last digit.
Second to last digit should be odd
It can be filled by 1,3,5,7,9
5 choices to fill second to last digit.
Rest two places of the four digit pin can be fill in 10 ways each
Total number of choices to form four digit pin
= 10 × 10 × 5 × 5
= 2,500
Therefore, number of choices for the given set of digits {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} to make a pin of four digit with the condition of last digit to be even and second to last must be odd is equal to 2,500.
Learn more about number of choices here
brainly.com/question/14655459
#SPJ4
