# Numbers with distinct digits from 1000-9999

## Homework Statement

A) how many numbers have distinct digits from 1000 to 9999?

B) how many odd numbers have distinct digits from 1000 to 9999?

## The Attempt at a Solution

a) the first place value has choices from 1-9, the second has choices from 0-9 but one number was used in the first place value so it has 9 choices, third place has 8 choices and fourth place has 7 choices so

9*9*8*7 numbers = 4536 numbers with distinct digits

b)

the available digits are 1,3,5,7,9

so first place has 5 choices, second has 4, third has 3, fourth has 2

so 5*4*3*2 numbers = 120 odd numbers with distinct digits.

that question seemed a little too easy so thats why I am posting it.

## Answers and Replies

Part (a) is correct, but (b) is not. You have solved the case where each digit of the number is odd, however, the question asks for odd numbers, which means only the last digit will be odd.

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Part (a) is correct, but (b) is not. You have solved the case where each digit of the number is odd, however, the question asks for odd numbers, which means only the last digit will be odd.

ah okay, that makes

sense so 9*9*8*5 numbers = 3240 numbers with distinct digits that are odd.

Not quite. You already decided to reserve an odd digit for the units place, and you cannot have zero in the first digit. How many possibilities does that leave for the first digit? Apply the argument further to the second and third digits.

so first will have 8 choices, second will have 8 because theres the reserved digit and the digit used in the first place but now zero is available, so third will have 7 and the fourth has 5

8*8*7*5 = 2240 numbers

Yep, that's it! 