Answer:76,875
Step-by-step explanation:Assuming you need to know the number of 5-digit numbers which are odd, plus the number of 5-digit numbers which are even and each containing at least 2 odd digits:Number of 5-digit odd numbers = 9*10*10*10*5 = 45,000.Number of 5-digit even numbers = 9*10*10*10*5 = 45,000.Therefore, to get the total we need to subtract from the even numbers all numbers which have zero or one odd digit in them, then add the rest to the 45,000 odd numbers.Number of 5-digit even number with no odd digit:We have 4 choices for the first digit, and 5 choices for the other 4 digits: Total = 4*5*5*5*5 = 2,500.Number of 5-digit even number with one odd digit:If the odd digit is in the first position, we have 5 choices for it, and 5 choices for each of the 4 even digits: Total = 5*5*5*5*5 = 3,125.If the odd digit is in position 2, 3, or 4, we have 4 choices for the first digit, and 5 choices for each of the other 4 digits : Total = 3 * 4 * 5 * 5 * 5 * 5 = 7,500.Therefore, total number of 5-digit even numbers having zero or one odd digit in them: 2,500 + 3,125 + 7,500 = 13,125.Therefore, total number of 5-digit even numbers having at least 2 odd digits: 45,000 - 13,125 = 31,875.Therefore, total number of 5-digit numbers which are odd or have at least 2 odd digits:45,000 + 31,875 = 76,875.Good luck!