Last visit was: 27 Apr 2026, 11:46 It is currently 27 Apr 2026, 11:46
Home
Messages
Tests
Schools
Events
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
   1   2 
805+ (Hard)|   Combinations|      
avatar
Tapesh03
avatar
Joined: 21 Jun 2013
Last visit: 08 Dec 2025
Posts: 28
Own Kudos:
110
Given Kudos: 83
Posts: 28
Kudos: 110
A rectangular floor measures 2 by 3 meters. There are 5 whit 25 Nov 2017, 18:38
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi,

please can you help why simple combination formula is wrong to apply in this situation. As there are 15 tiles and 6 spaces, I thought of applying 15C6 as I have to choose 6 tiles from 15 and it can be any colors and arrangement doesn't matter. kindly help me advise why my approach is wrong.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 27 Apr 2026
Posts: 109,928
Own Kudos:
811,595
Given Kudos: 105,914
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,928
Kudos: 811,595
A rectangular floor measures 2 by 3 meters. There are 5 whit 26 Nov 2017, 02:02
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Tapesh03
Hi,

please can you help why simple combination formula is wrong to apply in this situation. As there are 15 tiles and 6 spaces, I thought of applying 15C6 as I have to choose 6 tiles from 15 and it can be any colors and arrangement doesn't matter. kindly help me advise why my approach is wrong.
Show more

Not only arrangement of colors in 5 tiles matter but also the number of each color in 6 tiles. You should understand it better if you re-read the solutions above carefully.
User avatar
elPatron434
User avatar
BSchool Moderator
Joined: 22 Jul 2018
Last visit: 12 Jun 2024
Posts: 473
Own Kudos:
373
Given Kudos: 618
GRE 1: Q165 V163
GRE 2: Q165 V163
GRE 3: Q170 V163
WE:Engineering (Computer Software)
Products:
My Reviews
GRE 1: Q165 V163
GRE 2: Q165 V163
GRE 3: Q170 V163
Posts: 473
Kudos: 373
A rectangular floor measures 2 by 3 meters. There are 5 whit 25 Sep 2021, 06:49
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi Bunuel
Request your help with the shorter solution posted above.

We have 6 spaces for each of the three types of tiles. Now the first 5 tiles would have 3 ways to fill them. However, the last tile might have either 2 ways (if all the 5 tiles are filled by the same colour) or 3 ways (if the previous 5 have been filled by atleast 2 colours).

In this case how do we determine the number of ways in which the last tile can be filled?

Thank You
avatar
ChandrahasM
avatar
Joined: 10 May 2019
Last visit: 31 Oct 2022
Posts: 4
Own Kudos:
5
Given Kudos: 5
Location: United States
Concentration: Technology, Leadership
GRE 1: Q165 V154
WE:Design (Computer Software)
GRE 1: Q165 V154
Posts: 4
Kudos: 5
A rectangular floor measures 2 by 3 meters. There are 5 whit 06 Oct 2021, 20:28
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
A rectangular floor measures 2 by 3 meters. There are 5 white, 5 black, and 5 red parquet blocks available. If each block measures 1 by 1 meter, in how many different color patterns can the floor be parqueted?
(A) 104
(B) 213
(C) 577
(D) 705
(E) 726

There are 5 white, 5 black, and 5 red blocks available to fill 2*3=6 slots. Following 6 cases are possible for different pattern arrangements:

5-1: 5 blocks of the same color and 1 block of different color: \(C^1_3*C^1_2*\frac{6!}{5!}=36\), (where \(C^1_3\) is ways to choose 1 color from 3, which will provide us with 5 blocks, \(C^1_2\) is ways to choose 1 color from 2 colors left, which will provide us with 1 blocks, and \(\frac{6!}{5!}\) is ways of different arrangements of 6 blocks, XXXXXY, out of which 5 are identical);

4-2: 4 blocks of the same color and 2 block of different color: \(C^1_3*C^1_2*\frac{6!}{4!*2!}=90\), (where \(C^1_3\) is ways to choose 1 color from 3, which will provide us with 4 blocks, \(C^1_2\) is ways to choose 1 color from 2 colors left, which will provide us with 2 blocks, and \(\frac{6!}{4!*2!}\) is ways of different arrangements of 6 blocks, XXXXYY, out of which 4 X's and 2 Y's are identical);

The same way for other patterns:
4-1-1: \(C^1_3*\frac{6!}{4!}=90\);
3-3: \(C^2_3*\frac{6!}{3!*3!}=60\);
3-2-1: \(C^1_3*C^1_2*\frac{6!}{3!*2!}=360\);
2-2-2: \(\frac{6!}{2!*2*2!}=90\);

Total: 36+90+90+60+360+90=726.

Answer: E.

Shorter approach:
Imagine the case in which we have not 5 blocks of each color but 6, then each slot from 2*3=6 would have 3 color choices to be filled with: white, black, or red. That means that total different ways to fill 6 slots would be 3*3*3*3*3*3=3^6;

Now, what is the difference between this hypothetical case and the one in the question? As we allowed 6 blocks of each color instead of 5, then we would get 3 patterns which are impossible when we have 5 blocks of each color: all white, all red and all black. Thus we should subtract these 3 cases: 3^6-3=726.

Answer: E.

kt00381n
Why 15!/6!9! wont work? Im confused. :oops:
I you look at the first approach you'll see that \(C^6_{15}\) doesn't give all possible patterns possible. \(C^6_{15}\) is # of different groups of 6 possible out of 15 distinct objects, which clearly is not the case here.

diddygmat
This does not make sense. Are we not counting repeating patterns here? The question syas different pattern...

for example: The pattern BBBRRR is being counted at least a 2 times here. I s my understanding right?

Thanks!
Both approaches above count different patterns. Consider the following case: there are 2 slots to fill and 4 white, 4 black, and 4 red blocks available. How many different arrangements are possible:
WW
WB
BW
WR
RW
BB
BR
RB
RR

Total of 9 different arrangements. The same as if we consider approach #2 from above: each slot from 2 has 3 color choices to be filled with: white, black, or red. That means that total different ways to fill 2 slots would be 3*3=3^2=9.

Hope it helps.
Show more
Shorter approach is dope
avatar
vignesh2895
avatar
Joined: 22 Sep 2021
Last visit: 11 Jan 2022
Posts: 5
Own Kudos:
2
Given Kudos: 6
Location: India
Concentration: Statistics, Finance
WE:Analyst (Finance: Investment Banking)
Posts: 5
Kudos: 2
A rectangular floor measures 2 by 3 meters. There are 5 whit 06 Oct 2021, 22:32
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
A rectangular floor measures 2 by 3 meters. There are 5 white, 5 black, and 5 red parquet blocks available. If each block measures 1 by 1 meter, in how many different color patterns can the floor be parqueted?
(A) 104
(B) 213
(C) 577
(D) 705
(E) 726

There are 5 white, 5 black, and 5 red blocks available to fill 2*3=6 slots. Following 6 cases are possible for different pattern arrangements:

5-1: 5 blocks of the same color and 1 block of different color: \(C^1_3*C^1_2*\frac{6!}{5!}=36\), (where \(C^1_3\) is ways to choose 1 color from 3, which will provide us with 5 blocks, \(C^1_2\) is ways to choose 1 color from 2 colors left, which will provide us with 1 blocks, and \(\frac{6!}{5!}\) is ways of different arrangements of 6 blocks, XXXXXY, out of which 5 are identical);

4-2: 4 blocks of the same color and 2 block of different color: \(C^1_3*C^1_2*\frac{6!}{4!*2!}=90\), (where \(C^1_3\) is ways to choose 1 color from 3, which will provide us with 4 blocks, \(C^1_2\) is ways to choose 1 color from 2 colors left, which will provide us with 2 blocks, and \(\frac{6!}{4!*2!}\) is ways of different arrangements of 6 blocks, XXXXYY, out of which 4 X's and 2 Y's are identical);

The same way for other patterns:
4-1-1: \(C^1_3*\frac{6!}{4!}=90\);
3-3: \(C^2_3*\frac{6!}{3!*3!}=60\);
3-2-1: \(C^1_3*C^1_2*\frac{6!}{3!*2!}=360\);
2-2-2: \(\frac{6!}{2!*2*2!}=90\);

Total: 36+90+90+60+360+90=726.

Answer: E.

Shorter approach:
Imagine the case in which we have not 5 blocks of each color but 6, then each slot from 2*3=6 would have 3 color choices to be filled with: white, black, or red. That means that total different ways to fill 6 slots would be 3*3*3*3*3*3=3^6;

Now, what is the difference between this hypothetical case and the one in the question? As we allowed 6 blocks of each color instead of 5, then we would get 3 patterns which are impossible when we have 5 blocks of each color: all white, all red and all black. Thus we should subtract these 3 cases: 3^6-3=726.

Answer: E.

kt00381n
Why 15!/6!9! wont work? Im confused. :oops:
I you look at the first approach you'll see that \(C^6_{15}\) doesn't give all possible patterns possible. \(C^6_{15}\) is # of different groups of 6 possible out of 15 distinct objects, which clearly is not the case here.

diddygmat
This does not make sense. Are we not counting repeating patterns here? The question syas different pattern...

for example: The pattern BBBRRR is being counted at least a 2 times here. I s my understanding right?

Thanks!
Both approaches above count different patterns. Consider the following case: there are 2 slots to fill and 4 white, 4 black, and 4 red blocks available. How many different arrangements are possible:
WW
WB
BW
WR
RW
BB
BR
RB
RR

Total of 9 different arrangements. The same as if we consider approach #2 from above: each slot from 2 has 3 color choices to be filled with: white, black, or red. That means that total different ways to fill 2 slots would be 3*3=3^2=9.

Hope it helps.
Show more

Why not like below? In total 21

5W 1B 0R
5W 0B 1R
0W 5B 1R
0W 1B 5R
1W 5B 0R
1W 0B 5R
4W 2B 0R
4W 0B 2R
0W 4B 2R
0W 2B 4R
2W 0B 4R
2W 4B 0R
4W 1B 1R
1W 4B 1R
1W 1B 4R
3W 2B 1R
2W 3B 1R
2W 1B 3R
3W 1B 2R
1W 2B 3R
1W 3B 2R
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,986
Own Kudos:
1,119
Posts: 38,986
Kudos: 1,119
A rectangular floor measures 2 by 3 meters. There are 5 whit 06 Oct 2025, 03:24
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.
   1   2 
Moderators:
Bunuel
Math Expert
109928 posts
Krunaal
Tuck School Moderator
852 posts