Problem Solving

soni_pallavi wrote:Q1) How many 4-digit positive integers can be formed by using digits from 1 to 9 so that two digits are equal to each other and the remaining two are also equal to each other but different from the other two?

a)400
b)1728
c)108
d)216
e)432

Here's a slightly different approach:

Take the task of creating 4-digit integers and break it into stages.

Stage 1: Choose the 2 digits that you will be working with.
There are 9 digits and we must choose 2 of them.
Since the order of the digits we select does not matter, we can use combinations.
We can select 2 digits from 9 digits in 9C2 ways (= 36 ways)
In other words, we can complete stage 1 in (36 ways).

Stage 2: Arrange the 4 digits.
In how many ways can we arrange 2 pairs of identical numbers (e.g., 1, 1, 2, and 2)?
When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:

If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]

For this question, we have 4 letters, and 2 are alike and another 2 are alike.
So, the total number of arrangements is 4!/[(2)(2)] = 6
In other words, we can complete stage 2 in (6 ways).

By the Fundamental Counting Principle (FCP) we can complete the 2 stages (and thus create a 4-digit integer) in (36)(6) ways ([spoiler]= 216 ways[/spoiler])

Answer = D

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