Combinations Doubt; HELP

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soni_pallavi: Master | Next Rank: 500 Posts; Posts: 107; Joined: Tue Jun 05, 2012 3:25 amFollowed by: 13 members; Thanked: 1 times

by soni_pallavi » Sat Nov 17, 2012 5:21 am

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

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Sat Nov 17, 2012 7:53 am

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

Number of ways to select 2 digits from 9 choices = 9C2 = 36.

Let A and B be the two digits selected.
Any arrangement of AABB will form a viable integer.
The number of ways to arrange 4 elements = 4!.
But here there are two pairs of IDENTICAL elements: AA and BB.
When an arrangement includes IDENTICAL elements, we must divide by the number of ways to arrange the identical elements.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
The result is a reduction in the number of unique arrangements.
Since the number of ways to arrange each identical pair = 2!, we get:
Number of ways to arrange AABB = 4!/2!2! = 6.

To combine the options above, we multiply:
36*6 = 216.

The correct answer is D.

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Sat Nov 17, 2012 8:41 am

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

Brent Hanneson - Creator of GMATPrepNow.com

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