Night reader wrote:How many four-digit odd numbers do not use any digit more than once?
(A) 1728
(B) 2160
(C) 2240
(D) 2268
(E) 2520
Rahul's solution above is great. It applies something called the By the Fundamental Counting Principle (FCP). This technique is also called the Slot Method.
It's important to note, that when you apply this strategy, you should
always begin with the most restrictive cases
Here's what I mean.
We're going to take the task of building acceptable 4-digit numbers and break it into stages.
Our stages will consist of selecting the four digits needed. Among these stages, the most restrictive one is the units digit, since it must be odd. So, we'll begin there.
Stage 1: Select the units digit
Since the digit must be 1, 3, 5, 7, or 9, we can complete this stage in
5 ways.
From here, the 2nd most restrictive stage is selecting the thousands digit (since it cannot be zero). So . ..
Stage 2: Select the thousands digit
This digit cannot equal zero and it cannot be the same as the digit selected as the units digit.
This leaves 8 digits to choose from, so we can complete this stage in
8 ways.
At this point, the two remaining stages are equally restrictive, so we can complete either stage next.
Stage 3: Select the tens digit
This digit cannot equal either of the 2 digits already-selected digits.
This leaves 8 digits to choose from, so we can complete this stage in
8 ways.
Stage 4: Select the hundreds digit
This digit cannot equal either of the 3 digits already-selected digits.
This leaves 7 digits to choose from, so we can complete this stage in
7 ways.
By the Fundamental Counting Principle (FCP) we can complete all 4 stages (and thus build a 4-digit integer) in
(5)(8)(8)(7) ways ([spoiler]= 2240 ways[/spoiler])
Cheers,
Brent
Aside: For more information about the FCP, we have a free video on the subject:
https://www.gmatprepnow.com/module/gmat-counting?id=775
