Combinatorics

This topic has expert replies
Newbie | Next Rank: 10 Posts
Posts: 9
Joined: Tue Jun 02, 2009 10:14 am

Combinatorics

by hooray » Thu Aug 20, 2009 12:35 pm
In how many ways can 3-digit numbers be formed selecting 3 digits from 1, 1, 2, 3, 4?

1. 5P3/(2!)
2. 4P3
3. 4^3
4. 4P3+3C1*(3!/2!)
5. 60 × 3!

Senior | Next Rank: 100 Posts
Posts: 59
Joined: Mon Jul 13, 2009 8:58 pm
Thanked: 10 times
GMAT Score:710

by ankitns » Thu Aug 20, 2009 1:22 pm
The answer should be [spoiler]A -> 5P3 / (2!)[/spoiler]
We select 3 numbers from 5 possible selections. But since 1 is repeated twice, we divide it by 2!.

Hope it helps.
Attempt 1: 710, 92% (Q 42, 63%; V 44, 97%)
Attempt 2: Coming soon!

User avatar
Site Admin
Posts: 2567
Joined: Thu Jan 01, 2009 10:05 am
Thanked: 712 times
Followed by:550 members
GMAT Score:770

by DanaJ » Thu Aug 20, 2009 1:26 pm
You have 5 objects (1, 1, 2, 3, 4) from which you need to choose 3. The order matters, so for this you'll have the formula 5P3. However, since 1 appears 2 times, you'll need to reduce the whole thing by 2!. This is why I am getting A.

Newbie | Next Rank: 10 Posts
Posts: 9
Joined: Tue Jun 02, 2009 10:14 am

by hooray » Thu Aug 20, 2009 2:01 pm
I don't think that's right since it's not a pure combination. 112 is still a distinct number from 121, though if you are using the combinations method they will be treated as identical. I believe the OA is (D), I'm just not sure how they got it.

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 2621
Joined: Mon Jun 02, 2008 3:17 am
Location: Montreal
Thanked: 1090 times
Followed by:355 members
GMAT Score:780

by Ian Stewart » Thu Aug 20, 2009 4:10 pm
If you use three different digits, you'd have 4 choices for the hundreds digit, 3 for the tens, and 2 for the units, so 4*3*2 choices in total. If you use two 1's and a different digit (2, 3 or 4), you have three choices for which other digit to use, and three choices for where to put it - the hundreds, tens or units place - so 3*3 numbers you can make with two 1's and something else. So you have 4*3*2 + 3*3 = 24 + 9 = 33 different numbers in total.

You will *never* see the number 33 written as "4P3+3C1*(3!/2!)" in a GMAT question - it's not only an absurd way to write a number as small as 33, but notation like nCr or nPr is not something GMAT test takers are required to know about for the test (in mathematics, there are about a dozen different notations that are used for those quantities - notation like nCr is not standard in mathematics, and many people will have studied a different notation, which is perfectly fine). So I'm curious about the source of the question; the answer choices are not GMAT-like.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com

Newbie | Next Rank: 10 Posts
Posts: 9
Joined: Tue Jun 02, 2009 10:14 am

by hooray » Thu Aug 20, 2009 4:17 pm
Ahhh... thank you!

I'm not sure what the source of the question is but I'll ask. It was posted on a DS subforum of gmatclub... I take my GMAT in a couple of days and was sort of shocked since I've never seen this type of combinatorics questions before. Thanks for responding!