maybe is wrong, but as with 2 chairs out of 5 you get 10 possible combinations and the total combinations are 150, then the combinations of tables must be 15... only 6 gives this possibility.
it took me a while to get to this, any fast solution?
BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course
Redeem
Featured Course
Target Test Prep GMAT OnDemand
Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.
- 715+ score guarantee — highest in the industry (99th percentile)
- 52 chapters · 1,500+ lessons · 4,000+ practice questions
- 400 hours of video · 1,500+ instructor-led HD smartboard lessons
- 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
- 24/7 live support + weekly Zoom office hours with GMAT instructors
- TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
- 1,200+ flashcards + AI-powered study assistant & daily calendar
- OnDemand, LiveTeach & GMAT Bootcamp formats available
- Also: GRE, SAT Math & Executive Assessment courses
- MBA Admissions Consulting now available
- 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
- 200,000+ students served
- 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★
5.0
(559 reviews)
130-pt guarantee
$0 to start
then $127/mo
Chairs
Source: Beat The GMAT — Problem Solving |
Why cant the number of chairs selection be 20? P(5,2).
First chair can be selected in 5 ways and second can be selected in 4 ways.Therefore Total = 5*4=20!...
First chair can be selected in 5 ways and second can be selected in 4 ways.Therefore Total = 5*4=20!...
mmm i m a new, so i m not that confident, but i though:
it says there are 5 chairs, example: A B C D E.
possible selections
AB AC AD AE
BC BD BE
CD CE
DE
i think...
ps. by the way, ricky stands for riccardo? i m looking for someone in milan to study with...
it says there are 5 chairs, example: A B C D E.
possible selections
AB AC AD AE
BC BD BE
CD CE
DE
i think...
ps. by the way, ricky stands for riccardo? i m looking for someone in milan to study with...
Since a total of 150 combinations are there. It has to be
150=No. of combinations of chairs* no. of combinations of tables
No. of ways chairs can be arranged=5C2=10
Therefore, no. of ways tables can be arranged would be 15.
6C2=15. Hence, there has to be 6 tables.
150=No. of combinations of chairs* no. of combinations of tables
No. of ways chairs can be arranged=5C2=10
Therefore, no. of ways tables can be arranged would be 15.
6C2=15. Hence, there has to be 6 tables.
I understand everything but why cant be there 20 possible selections of chairs.
First chair can be selected in 5 ways.Second in 4 ways.So why its not 5*4=20 ?
First chair can be selected in 5 ways.Second in 4 ways.So why its not 5*4=20 ?
ricky wrote:I understand everything but why cant be there 20 possible selections of chairs.
First chair can be selected in 5 ways.Second in 4 ways.So why its not 5*4=20 ?
first chair (A) can be selected with 4 other chairs: AB AC AD AE
second chair (B) can then be selected in 3 other chairs BC BD BE
third chair (c) can be selected in 2 more ways: CD CE
last two chairs can be selected togher: DE
4+3+2+1=10
I think the answer is 6 it can never be 30
no. of chairs = 5
no. of tables = x
5C2 * xC2 = 150
10 * xC2 = 150
xC2 = 15
6C2 = 15
Hence the answer is 6.
GMAT prep software is known for having bugs such as these.
no. of chairs = 5
no. of tables = x
5C2 * xC2 = 150
10 * xC2 = 150
xC2 = 15
6C2 = 15
Hence the answer is 6.
GMAT prep software is known for having bugs such as these.
Please See the question does NOT mention if their chairs and Tables are unique to each other. Hence, we can NOT name chairs as A,B,C,D..
& 5C2 * NC2 = 150.
=> N =6
& 5C2 * NC2 = 150.
=> N =6
-
Stuart@KaplanGMAT
- GMAT Instructor
- Posts: 3225
- Location: Toronto
- Thanked: 1710 times
- GMAT Score:800
Because you have duplicates - the way you counted, chair A then chair B is a different selection than chair B then chair A - when in fact those selections are identical.ricky wrote:I understand everything but why cant be there 20 possible selections of chairs.
First chair can be selected in 5 ways.Second in 4 ways.So why its not 5*4=20 ?
So, you could do the calculation (5*4)/2 = 20/2 = 10
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
HI Stuart,
Please can you explain the concept of deplicates in per./com./prob. with some example.
Thanks in advance
Please can you explain the concept of deplicates in per./com./prob. with some example.
Thanks in advance
-
Stuart@KaplanGMAT
- GMAT Instructor
- Posts: 3225
- Location: Toronto
- Thanked: 1710 times
- GMAT Score:800
Duplicates occur when you count the same arrangement more than once.arorag wrote:HI Stuart,
Please can you explain the concept of deplicates in per./com./prob. with some example.
Thanks in advance
In combinations, this happens when we forget that order doesn't matter.
For example, let's say I want to choose a pair of people out of 4 possibilities, A B C and D.
If I were to say "well, there's 4 possibilities for the first person and 3 possibilities for the second person, so there's 4*3 = 12 possible pairs", I'd be double counting because I'm forgetting that the pair "AB" is the exact same as the pair "BA".
In permutations, duplicates happen when we have identical entities, i.e. not all the entities are unique.
For example, if I want to know all the possible ways to arrange the letters in the word "GMAT", the answer would simply be 4!.
However, if I want to know all the possible ways to arrange the letters in the word "KAPLAN" and use the same approach, I'd get an answer of 6!, which would be wrong, since "KAPLAN" has two As.
Let's call the As in Kaplan A(1) and A(2) and see what I mean.
KA(1)PLA(2)N and KA(2)PLA(1)N look exactly the same, but we've counted it as two separate arrangements.
In order to avoid duplications in permutation questions, you need to factor out the number of duplicates.
So, the true number of arrangements of "KAPLAN" is 6!/2!.
Let's look at some other examples:
DESSERT... 7!/2!2!, since we need to factor out 2 Es and 2 Ss.
DESSERTS... 8!/2!3!, since we need to factor out 2 Es and 3 Ss.
DESERTERS... 9!/3!2!2!, since we need to factor out 3 Es, 2 Ss and 2 Rs.
The basic formula is n!/r!s!t!... in which n is the total number of items and r, s, t, ... are the number of duplications.
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
