Matrix

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santhoshsram: Senior | Next Rank: 100 Posts; Posts: 62; Joined: Wed Oct 26, 2011 2:30 pm; Thanked: 7 times

by santhoshsram » Tue Jan 10, 2012 10:15 pm

In how many different ways can 3 As, 3Bs and 3Cs be arranged in a 3x3 matrix so that no row or column has more than one of each of the alphabets?

-- Santhosh S

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Source: Beat The GMAT — Problem Solving | Tue Jan 10, 2012 10:15 pm

[email protected]: Senior | Next Rank: 100 Posts; Posts: 57; Joined: Mon Dec 21, 2009 6:27 am; Location: Melbourne, Australia; Thanked: 17 times

by [email protected] » Wed Jan 11, 2012 12:06 am

I'm not quite sure abouth this one.
1st row can be formed in 3x3x3 ways and to make their position fixed, they can be arranged in 3 ways.So total is 81.
Let's say the 1st row looks like this now
A B C
X X X
X X X

Now 1st column can be chosen in 2x2 (from 2B & 2C)=4 ways and themselves cna be arranged in 2 ways. So total is 81*8.

A B C
B X X
C X X

2nd column can be chosen in 1X2 ( 1C & 2A) ways and among themselve 2 ways. So total is 81*8*4

A B C
B C X
C A X

3rd column can be chosen in 1x1 (1A & 1B) ways and among themselve in 2 ways,

A B C
B C A
C A B
Total is 81*8*4*2=5184 (Seems to big!)

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santhoshsram: Senior | Next Rank: 100 Posts; Posts: 62; Joined: Wed Oct 26, 2011 2:30 pm; Thanked: 7 times

by santhoshsram » Wed Jan 11, 2012 12:34 am

Way too big indeed

Are you sure there are 81 ways to form the first row?

[email protected] wrote:I'm not quite sure abouth this one.
1st row can be formed in 3x3x3 ways and to make their position fixed, they can be arranged in 3 ways.So total is 81.
Let's say the 1st row looks like this now
A B C
X X X
X X X

Now 1st column can be chosen in 2x2 (from 2B & 2C)=4 ways and themselves cna be arranged in 2 ways. So total is 81*8.

A B C
B X X
C X X

2nd column can be chosen in 1X2 ( 1C & 2A) ways and among themselve 2 ways. So total is 81*8*4

A B C
B C X
C A X

3rd column can be chosen in 1x1 (1A & 1B) ways and among themselve in 2 ways,

A B C
B C A
C A B
Total is 81*8*4*2=5184 (Seems to big!)

-- Santhosh S

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[email protected]: Senior | Next Rank: 100 Posts; Posts: 57; Joined: Mon Dec 21, 2009 6:27 am; Location: Melbourne, Australia; Thanked: 17 times

by [email protected] » Wed Jan 11, 2012 1:46 am

I'm bit confused in this problem.
If we consider all the A,B & C alphabets a a unit then 1st row can be formed in:
3C1*3+2C1*3+1C1*3 ways..
Am i wrong again?

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Neo Anderson: Senior | Next Rank: 100 Posts; Posts: 92; Joined: Thu Oct 06, 2011 8:06 am; Thanked: 18 times

by Neo Anderson » Wed Jan 11, 2012 6:21 am

In how many different ways can 3 As, 3Bs and 3Cs be arranged in a 3x3 matrix so that no row or column has more than one of each of the alphabets?

1) Choose first A to place, you have 9 places to choose: 9 ways
2) choose second A to place , you have 4 places to choose : 4 ways
3) choose 3rd A to place , you have only 1 place to choose : 1 way

Thus 9 X 4 X 1 ways for all A's

similarly for B and C

6 X 3 X 1 and 3 X 2 X 1 ways

Thus total ways: 9 X 4 X 6 X 3 X 3 X 2
also All As, Bs and Cs can be interchanged in their position thus we need to divide this by 3! three times.
that gives us 18 as answer!

correct me if I'm wrong!

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user123321: Master | Next Rank: 500 Posts; Posts: 385; Joined: Fri Sep 23, 2011 9:02 pmFollowed by: 6 members; Thanked: 62 times

by user123321 » Wed Jan 11, 2012 6:27 am

santhoshsram wrote:In how many different ways can 3 As, 3Bs and 3Cs be arranged in a 3x3 matrix so that no row or column has more than one of each of the alphabets?

it will be difficult to answer without choices.
it sometimes allows us to ballpark few choices.
IMO [spoiler]36?[/spoiler]

user123321

Just started my preparation

Want to do it right the first time.

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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 » Wed Jan 11, 2012 7:52 am

santhoshsram wrote:In how many different ways can 3 As, 3Bs and 3Cs be arranged in a 3x3 matrix so that no row or column has more than one of each of the alphabets?

Top row:
A, B and C must each appear once in the top row.
Number of ways to arrange A, B and C = 3! = 6.

Second row:
Number of options for A = 2. (Either of the 2 columns not occupied by A in the top row.)
Once A is placed in the second row, there is only ONE acceptable arrangement for B and C in the second row.

To illustrate:
ABC
XAX

implies

ABC
CAB

Thus, once A has been placed, the number of options for B and C = 1.
To combine our options for the 2nd row, we multiply:
2*1 = 2.

Bottom row:
Once A, B and C are placed in the top two rows, there is only ONE acceptable arrangement for the bottom row.

To illustrate:
ABC
CAB

implies

ABC
CAB
BCA

Thus, the number of options for the bottom row = 1.

Total number of possible arrangements:
To combine our options for each row, we multiply:
6*2*1 = 12.

An alternate approach would be to write out all the possible options:

ABC...ABC...ACB...ACB
CAB...BCA...BAC...CBA
BCA...CAB...CBA...BAC

BAC...BAC...CAB...CAB
ACB...CBA...ABC...BCA
CBA...ACB...BCA...ABC

BCA...BCA...CBA...CBA
ABC...CAB...ACB...BAC
CAB...ABC...BAC...ACB

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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
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I unlock the best way for YOU to solve problems.

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santhoshsram: Senior | Next Rank: 100 Posts; Posts: 62; Joined: Wed Oct 26, 2011 2:30 pm; Thanked: 7 times

by santhoshsram » Wed Jan 11, 2012 11:56 am

Crystal clear explanation from Mitch, as always. Thanks Mitch.

-- Santhosh S

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ArunangsuSahu: Master | Next Rank: 500 Posts; Posts: 382; Joined: Thu Mar 31, 2011 5:47 pm; Thanked: 15 times

by ArunangsuSahu » Thu Jan 12, 2012 7:11 am

CLUE: For each COLUMN and ROW the other two COLUMNS or ROWS can happen in 2 ways.

So for Column: 2+2+2=6 ways
and for Rows: = 2+2+2=6

Total=6+6=12 ways

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