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I'm having troubles with the "multiplication rule" which states:

If an operation consist of k steps and

the first step can be performed in n_1 ways,

the second step can be performed in n_2 ways [regardless of how the first step was peformed],

.

.

.

the kth step can be performed in n_k ways [regardless of how the preceding seps were performed],

then the entire operation can be performed in (n_1)(n_2)...(n_k) ways.

The book does an example with a simliar problem, but small numbers, the book did the following:

10 to 99

They found out the following:

How many integers from 10 to 99?

answer: (number of ways to pick the first digit)(number of ways to pick the 2nd digit) = (9)(10) = 90.

How many odd integers from 10 to 99?

Well odd integers are: 1,3,5,7,9....a total of 5.

answer: (number of ways to pick the first digit)(number of ways to pick the 2nd digit) = (9)(5) = 45.

Number of integers with distinct digits:

answer: (number of ways to pick the first digit)(number of ways to pick the 2nd digit) = (9)(9) = 81.

Number of odd integers with distinct digits:= (5)(8) = 40.

answer: (number of ways to pick the 2nd digit)((number of ways to pick the first digit)

First digit can't be 0, nor can it equal the 2nd digit.

Okay so i'm trying to apply that method to the bigger problem....

1000 to 9999

So for the number of odd integers with distinct digits...could i do the following:

(# of ways to pick 4th digit)(# of ways to pick 3rd digit)(# of ways to pick 2nd digit)(# of ways to pick 1st digit) = (5)(8)(7)(6) = 1680 ?

Thanks!

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# Homework Help: How many odd integers from 1000 to 9999 have distint digits? will this method work?

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