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

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
Vote for Target Test Prep, Newsweek Readers’ Choice Awards 2026
NEWSWEEK READERS’ CHOICE 2026

BIG NEWS! Target Test Prep has been nominated, and they’d love your vote!

TTP has worked incredibly hard to build the best test prep experience possible, and winning Newsweek’s 2026 Readers’ Choice Award for Best Test Prep would mean a lot to them. If TTP has helped you, they’d be incredibly grateful for your vote. You can vote once each day through September 9.

Vote for TTP
GMATLiveTeach 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions
Start free 5-day trial
Read 559 reviewsTarget Test Prep

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Combinations Randolph Question - Need expert help

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 405
Joined: Thu Feb 10, 2011 1:44 amFollowed by: 1 member
Thanked: 3 times
by voodoo_child » Sun Jul 29, 2012 2:44 pm
Randolph has a deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, there are 2 cards in the deck that have the same value.

Randolph likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for a pair of cards that have the same value. How many such combinations are possible?

A. 240
B. 960
C. 120
D. 40
E. 5760


I tried the above problem using Combinations, and I got the correct answer which is A.

Here's the method : 2* (6C1 * 1C1 * 5C2) + (6C2 * 2C1 * 4C1) = 240 {First part - Choose 1 card from the first suit, remaining two cards from the second suit; Second part - Choose 2 from each} {1C1 and 2c1 represent the same numbered card}


However, I crashed while using Permutations. Permutations - 12*1*10*8 = 960. (crash ) I don't know how I can arrive at 240 using this method. Really confused.

I tried a couple of other examples:

LEt's say there are only 4 cards of two suits each.

Combinations: 2* (4C1*1C1*3C1) + (4C2*2C1*2C1) = 48 {Same logic - choose 1 from the first suit; remainign three from the second suit; Second part - Choose 2 from each. }
Using permutations - 8* 1*6*4= 48*4...(crash )


I see that in both the examples, the "Combinations" number is "Permutations/4"...Is there a rule to quickly relate Permutations with Combinations?

I know that nCr = nPr/r!; However, I believe that this theorem is not applicable here.

Correct?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 126
Joined: Sun Jun 24, 2012 10:11 amFollowed by: 7 members
Location: Chicago, IL
Thanked: 36 times

by tutorphd » Sun Jul 29, 2012 3:18 pm
Your calculation of 12*1*10*8 = 960 permutations only includes the scenario in which the first two cards form a pair, and the other two don't.

There are additional scenarios. If P denotes a pair card and N denots a non-pair card, there are 6 different scenarios (anagrams of PPNN):

PPNN
NPPN
NNPP
PNNP
PNPN
NPNP

Each scenario has 960 permuations in it for a total of 6*960 permutations that have two cards forming a pair and two cards not forming a pair.

Since all the cards are different (either by value or by suit), the theorem you mentioned does apply and 4!=24 permutations correspond to a single combination.

The number of combinations is then 6*960/24 = 960/4 = 240.
Last edited by tutorphd on Sun Jul 29, 2012 3:44 pm, edited 1 time in total.
Skype / Chicago quant tutor in GMAT / GRE
https://gmat.tutorchicago.org/
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 126
Joined: Sun Jun 24, 2012 10:11 amFollowed by: 7 members
Location: Chicago, IL
Thanked: 36 times

by tutorphd » Sun Jul 29, 2012 3:29 pm
The most efficient approach to this problem is neither combinations nor permutations but using the slot principle (multiply possibilities for each slot):

total combinations = (ways to choose two pair cards) x (ways to choose two non-pair cards)

ways to choose two pair cards = 6 because there are 6 pairs total when the order of the cards in the pair does not matter

ways to choose two non-paired cards = 10*8/2!=40 because 10 cards are left after a pair is chosen and then 8 cards are left that won't match the first chosen non-pair cards. The factorial 2! corrects for the fact that the card order does not matter in the non-pair.

So total number of combinations is 6*40 = 240.
Skype / Chicago quant tutor in GMAT / GRE
https://gmat.tutorchicago.org/
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 405
Joined: Thu Feb 10, 2011 1:44 amFollowed by: 1 member
Thanked: 3 times

by voodoo_child » Sun Jul 29, 2012 4:22 pm
tutorphd wrote:Your calculation of 12*1*10*8 = 960 permutations only includes the scenario in which the first two cards form a pair, and the other two don't.

There are additional scenarios. If P denotes a pair card and N denots a non-pair card, there are 6 different scenarios (anagrams of PPNN):

PPNN
NPPN
NNPP
PNNP
PNPN
NPNP

Each scenario has 960 permuations in it for a total of 6*960 permutations that have two cards forming a pair and two cards not forming a pair.

Since all the cards are different (either by value or by suit), the theorem you mentioned does apply and 4!=24 permutations correspond to a single combination.

The number of combinations is then 6*960/24 = 960/4 = 240.
Awesome. Can you please explain why you did "Since all the cards are different (either by value or by suit), the theorem you mentioned does apply and 4!=24 permutations correspond to a single combination."
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 126
Joined: Sun Jun 24, 2012 10:11 amFollowed by: 7 members
Location: Chicago, IL
Thanked: 36 times

by tutorphd » Sun Jul 29, 2012 5:02 pm
The theorem applies only when you are selecting from a set of different objects. If some of the objects are not different/distinguishable, not all combinations will correspond to the same number of permutations, and the number of permutations and combinations are not related so easily by a simple division.

Take for example 3 throws of a fair coin. The possible outcomes are 8 permutations, like HTH etc.

The combination {H, T, H} corresponds to 3 permutations: THH, HTH, HHT.
The combination {H, H, H} corresponds to just 1 permutation: HHH.
So you can't find all combinations just by dividing all permutations by 3!.
Skype / Chicago quant tutor in GMAT / GRE
https://gmat.tutorchicago.org/
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 405
Joined: Thu Feb 10, 2011 1:44 amFollowed by: 1 member
Thanked: 3 times

by voodoo_child » Tue Jul 31, 2012 5:28 am
Tutorphd,
Thanks for your analysis. However, I didn't follow this part "Since all the cards are different (either by value or by suit) and 4!=24 permutations correspond to a single combination. " in your post.

There are four cards, out of which two of them are same (=pair). Hence, a combination could look like AABC => there are 4!/2! permutations and not 4!. Why did you divide by 24?

Thoughts please?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 126
Joined: Sun Jun 24, 2012 10:11 amFollowed by: 7 members
Location: Chicago, IL
Thanked: 36 times

by tutorphd » Tue Jul 31, 2012 7:47 pm
Because the permutation calculation of 960 favorable permutations in each scenario treats each card as distinguishable by number or suit or both - that is implicit in the calculation used to get 960. You don't correct there for 2 in suit A and 2 of suit B "being the same".

That is why, when you convert the number of permutations to number of combinations, you have to keep treating all the cards as different. For example, if the two suits are denoted by A and B, the combination {2A, 2B, 1A, 3A} corresponds to 24 permutations of the 4 different elements.
Skype / Chicago quant tutor in GMAT / GRE
https://gmat.tutorchicago.org/
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 15
Joined: Wed Jul 25, 2012 9:02 am
Thanked: 1 times

by armand_h » Wed Aug 01, 2012 12:30 am
you can also do the following:
with 12 cards you can form 6 pairs of identical cards: P1, P2, P3, P4, P5, P6
Once you have one pair of identical cards, you are left with 10 cards made up of only 2 suits of 5 cards each.
with these 10 cards you can make 45 combinations of 2 cards 10C2=45.
The 45 combinations contain 5 pairs of identical cards and 40 pairs of different cards.
In order to have one pair in a set of 4 cards, you need to pick one pair from 6 and one pair from 40 (one pair from the identical cards, and one pair from the non identical cards)

Total number of combinations[spoiler]=6*40=240[/spoiler][/spoiler]
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 405
Joined: Thu Feb 10, 2011 1:44 amFollowed by: 1 member
Thanked: 3 times

by voodoo_child » Wed Aug 01, 2012 8:04 am
tutorphd wrote:Because the permutation calculation of 960 favorable permutations in each scenario treats each card as distinguishable by number or suit or both - that is implicit in the calculation used to get 960. You don't correct there for 2 in suit A and 2 of suit B "being the same".

That is why, when you convert the number of permutations to number of combinations, you have to keep treating all the cards as different. For example, if the two suits are denoted by A and B, the combination {2A, 2B, 1A, 3A} corresponds to 24 permutations of the 4 different elements.
I didn't folllow this. In calculating 960, we did "12*1*10*8" instead of "12*11*10*9" - why are you saying that the cards were assumed distinguishable? Only three cards were assumed distinguishable. I feel that I am missing something very fundamental....

Thoughts / help?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1