BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem
TTP OnDemand GMAT Video Course
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
Start Free 5-Day Trial Read 559 Reviews →

Combinations

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Legendary Member
Posts: 966
Joined: Sat Jan 02, 2010 8:06 amFollowed by: 21 members
Thanked: 230 times
by shankar.ashwin » Mon Oct 10, 2011 9:54 pm
There are 4 pens each of a different color and four caps each of a diff color (But same as that of the pens). In how many ways can these caps be put on the pens such that the pen and the cap do not have the same color.

A) 9
B) 12
C) 18
D) 23
E) 27
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
User avatar
Senior | Next Rank: 100 Posts
Posts: 68
Joined: Fri Dec 03, 2010 10:29 pmFollowed by: 1 member
Thanked: 5 times

by Bek » Mon Oct 10, 2011 10:24 pm
IMO: 12
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 23
Joined: Tue Oct 04, 2011 9:10 pm

by Proleefeek111 » Mon Oct 10, 2011 10:27 pm
shankar.ashwin wrote:There are 4 pens each of a different color and four caps each of a diff color (But same as that of the pens). In how many ways can these caps be put on the pens such that the pen and the cap do not have the same color.

A) 9
B) 12
C) 18
D) 23
E) 27
Pens: A, B, C, D <Colors depicted by alphabets>
Caps: a, b, c, d <small case letters depicting equivalent colors of the Pen >
Comboes allowed: Ab, Ac, Ad, Ba, Bc, Bd, Ca, Cb, Cd, Da, Db, Dc = 12
Ans > B
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 966
Joined: Sat Jan 02, 2010 8:06 amFollowed by: 21 members
Thanked: 230 times

by shankar.ashwin » Mon Oct 10, 2011 10:52 pm
Proleefeek111 wrote: Pens: A, B, C, D <Colors depicted by alphabets>
Caps: a, b, c, d <small case letters depicting equivalent colors of the Pen >
Comboes allowed: Ab, Ac, Ad, Ba, Bc, Bd, Ca, Cb, Cd, Da, Db, Dc = 12
Ans > B
12 is incorrect.

Try listing down pairs as (Ab)(Bc)(Cd)(Da)
where ABCD are pens and abcd are caps.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 504
Joined: Tue Apr 19, 2011 1:40 pmFollowed by: 11 members
Thanked: 114 times

by knight247 » Mon Oct 10, 2011 11:09 pm
The answer is A.. This is similar to that Tanya mailed 4 letters problem.
Look at the following link.

https://www.beatthegmat.com/difficult-pr ... 77390.html

That question is divided in two parts. The second part is the one relevant to this problem. Especially look at Anurag's explanation. And also one guy has listed the DeRangement formula which can help you find this answer in 15 seconds. lol
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 23
Joined: Tue Oct 04, 2011 9:10 pm

by Proleefeek111 » Tue Oct 11, 2011 12:37 am
shankar.ashwin wrote:
Proleefeek111 wrote: Pens: A, B, C, D <Colors depicted by alphabets>
Caps: a, b, c, d <small case letters depicting equivalent colors of the Pen >
Comboes allowed: Ab, Ac, Ad, Ba, Bc, Bd, Ca, Cb, Cd, Da, Db, Dc = 12
Ans > B
12 is incorrect.

Try listing down pairs as (Ab)(Bc)(Cd)(Da)
where ABCD are pens and abcd are caps.
Shouldn't the problem statement be changed as follows then:

"In how many ways can these pens be arranged such that a pen and it's cap do not have the same color. "

as opposed to

"In how many ways can these caps be put on the pens such that the pen and the cap do not have the same color. "
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
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 » Tue Oct 11, 2011 4:34 am
shankar.ashwin wrote:There are 4 pens each of a different color and four caps each of a diff color (But same as that of the pens). In how many ways can these caps be put on the pens such that the pen and the cap do not have the same color.

A) 9
B) 12
C) 18
D) 23
E) 27
Number of ways for NO pen to receive the correct cap = total ways to arrange the 4 caps - number of ways for 1, 2 or all 4 pens each to receive the correct cap.

Note that if 3 pens each receive the correct cap, then ALL 4 pens each receive the correct cap.

Let the 4 pens = ABCD and the 4 caps = abcd.

Total number of ways to arrange the 4 caps = 4! = 24.

Exactly 1 pen receives the correct cap:
Let A be the one pen to receive the correct cap.
Number of options for A = 1. (Must be a)
Number of options for B = 2. (Must be c or d)
At this point, either C or D could still receive the correct cap.
Number of options for the pen that could still receive the correct cap = 1. (Must be the WRONG cap)
Number of options for the last pen = 1. (Must be the one remaining cap)
To combine these options, we multiply:
1*2*1*1 = 2.
Since any of the 4 pens could be the one to receive the correct cap, we multiply by 4:
2*4 = 8.

Exactly 2 pens each receive the correct cap:
Let A and B be the pair to receive the corrrect caps.
Number of options for A and B = 1. (A must receive a, B must receive b)
Number of options for C = 1. (Must be d)
Number of options for D = 1 . (Must be c)
To combine these options, we multiply:
1*1*1 = 1.
Since any combination of 2 pens could be the pair to receive the correct caps, we multiply by the number of combinations of 2 that can be formed from 4 choices:
1 * 4C2 = 6.

All 4 pens each receive the correct cap:
Number of options = 1.

Thus:
Number of ways for NO pen to receive the correct cap = 24-8-6-1 = 9.

The correct answer is A.
Last edited by GMATGuruNY on Tue Oct 11, 2011 9:02 pm, edited 1 time in total.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
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.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Senior | Next Rank: 100 Posts
Posts: 68
Joined: Fri Dec 03, 2010 10:29 pmFollowed by: 1 member
Thanked: 5 times

by Bek » Tue Oct 11, 2011 7:40 am
4 pens ABCD
4 cups abcd

A B C D

b a d c
b d a c
b c d a

c a d b
c d b a
c d a b

d a b c
d b a c
d c a c
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 23
Joined: Tue Oct 04, 2011 9:10 pm

by Proleefeek111 » Tue Oct 11, 2011 8:58 pm
Bek wrote:4 pens ABCD
4 cups abcd

A B C D

b a d c
b d a c
b c d a

c a d b
c d b a
c d a b

d a b c
d b a c --> B cannot couple B
d c a c--> A typo, i think u meant d c a b
Mayte, check the comboes :).
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Senior | Next Rank: 100 Posts
Posts: 90
Joined: Fri Feb 25, 2011 12:56 amFollowed by: 3 members
Thanked: 1 times
GMAT Score:770

by sohrabkalra » Wed Oct 12, 2011 3:09 am
is there a way for this to be solved by directly counting ways of allocating caps to pens(other than listing all combinations) rather than counting total and then subtracting unwanted cases?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 1239
Joined: Tue Apr 26, 2011 6:25 amFollowed by: 26 members
Thanked: 233 times
GMAT Score:680

by sam2304 » Wed Oct 12, 2011 3:19 am
IMO B.

1 Pen 3 other colors. So for each colored pen we have 3 combinations.
4 x 3 = 12 combinations totally
Getting defeated is just a temporary notion, giving it up is what makes it permanent.
https://gmatandbeyond.blogspot.in/
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
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 Oct 12, 2011 5:16 am
sohrabkalra wrote:is there a way for this to be solved by directly counting ways of allocating caps to pens(other than listing all combinations) rather than counting total and then subtracting unwanted cases?
Let the pens = ABCD and the caps = abcd.

Case 1: The pens are divided into pairs. Within each pair, the pens swap caps.
Number of options for A = 3. (Must receive b, c or d)
Number of options for the pen whose cap was given to A = 1. (Must receive a)
2 pens left. Each could still receive the correct cap, since the first 2 pens swapped caps.
Number of options for the next pen = 1. (Of the 2 remaining caps, must receive the wrong cap)
Number of options for the last pen = 1. (Only 1 cap left)
To combine these options, we multiply:
3*1*1*1 = 3.

Case 2: No two pens swap caps.
Number of options for A = 3. (Must receive b, c, or d)
Number of options for the pen whose cap was given to A = 2. (Of the 3 remaining caps, either of the 2 caps besides a)
2 pens left. Only one could still receive the correct cap, since the first two pens did not swap caps.
Number of options for the pen that could still get the correct cap = 1. (Must receive the wrong cap)
Number of options for the last pen = 1. (Only 1 cap left)
To combine these options, we multiply:
3*2*1*1 = 6.

Total ways = 3+6 = 9.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
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.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1