Using the digits 0, 1, 2, ...8, 9, determine how many 4-digit numbers can be constructed according to the following criteria. The number must be odd and greater than 6,000; digits may be repeated.
The number of 4-digit numbers that can be constructed is

2000

Step-by-step explanation:

Since the number must be odd, it can have either 1, 3, 5, 7 or 9 in its one's place.

With a 4-digit number starting with 6, any of these digits can be in the one's place and it will still be greater than 6,000.

For the one's place, since we can choose from 5 digits, we have ⁵P₁. In the thousands place, we have one digit which is 6. For the tens and hundreds, place, we can choose from ten digits. So, we have ¹⁰P₁ and ¹⁰P₁ respectively.

So, the total number of 4 digit odd numbers greater than 6000 starting with 6 are 1 × ¹⁰P₁ × ¹⁰P₁ × ⁵P₁ = 1 × 10 × 10 × 5 = 500 numbers.

For the thousands place, we are left with the digits 7, 8 and 9. So, to arrange these 3 digits in the thousands place, we have ³P₁. For the tens and hundreds, place, we can choose from ten digits. So, we have ¹⁰P₁ and ¹⁰P₁ respectively. For the one's place, since we can choose from 5 digits, we have ⁵P₁.

So, the total number of 4 digit odd numbers greater than 6000 starting with 7, 8 or 9 are  ³P₁ × ¹⁰P₁ × ¹⁰P₁ × ⁵P₁ = 3 × 10 × 10 × 5 = 1500 numbers.

So, the total number of 4-digit numbers greater than 6000 that can be constructed is 500 + 1500 = 2000 numbers.

step-by-step explanation:

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about the problem:

srry i'm not done ! i'll