Last visit was: 29 Apr 2026, 23:41

It is currently 29 Apr 2026, 23:41

Toolkit

Practice Tests

Quantitative Questions

Problem Solving (PS)

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

Request Expert Reply

Confirm Cancel

Back to Forum

Create Topic

At a blind taste competition a contestant is offered 3 cups of each of

1 2 3

Zett1626

Joined: 04 Feb 2024

Last visit: 12 Dec 2025

Posts: 2

Own Kudos:

Given Kudos: 61

Posts: 2

Kudos: 1

25 Mar 2025, 20:50

Kudos

Bookmarks

Bunuel: please let me know where I go wrong.
I am not understanding where I go wrong with my work? Could you please help.

1- P(All 3)

Ways for all 3 is pick 2 for one type
ABCC
ABBC
AABC

Ways of arranging one example
4!/2! = 12

3 examples total - 3(12) = 36
Total ways 9C4 = 126

36/126 = 4/14 = P(All 3)

1- (4/14) = 10/14

Bunuel

In a blind taste competition, there are 3 types of tea: Type A, Type B, and Type C. Each type is presented in 3 cups, for a total of 9 cups. If a contestant tastes 4 different cups of tea at random, what is the probability that the contestant does not taste all 3 types of tea?

A. $\frac{1}{12}$

B. $\frac{5}{14}$

C. $\frac{4}{9}$

D. $\frac{1}{2}$

E. $\frac{2}{3}$

Experience GMAT Club Test Questions

Yes, you've landed on a GMAT Club Tests question

Craving more? Unlock our full suite of GMAT Club Tests here

Want to experience more? Get a taste of our tests with our free trial today

Rise to the challenge with GMAT Club Tests. Happy practicing!

GMAT Club Tests Check Our Products Try Free Tests

Pablopikachu

Joined: 05 Nov 2024

Last visit: 03 Sep 2025

Posts: 31

Own Kudos:

Given Kudos: 314

Posts: 31

Kudos: 12

18 Aug 2025, 12:31

Kudos

Bookmarks

Hi experts

In combinations and probability questions, when does order of selection actually matter and when not? How to figure that out? I end up miscalculating in such questions cuz I get confused whether to account for the order of selection or not. Cud someone clear this? Bunuel chetan2u KarishmaB

Krunaal

Tuck School Moderator

Joined: 15 Feb 2021

Last visit: 25 Apr 2026

Posts: 852

Own Kudos:

915

Given Kudos: 251

Status:Under the Square and Compass

Location: India

Schools: Tuck '26 (WL) Ross '26 (A$$$$) Stern '26 (A) McCombs '26 (A$)

GMAT Focus 1: 755 Q90 V90 DI82 Verified score

GPA: 5.78

WE:Marketing (Consulting)

Products:

Schools: Tuck '26 (WL) Ross '26 (A$$$$) Stern '26 (A) McCombs '26 (A$)

GMAT Focus 1: 755 Q90 V90 DI82 Verified score

Posts: 852

Kudos: 915

18 Aug 2025, 14:42

Kudos

Bookmarks

Pablopikachu

It will differ question to question, but broadly imo it is always good to ask after reading the question whether am I just selecting or am I arranging, if you're just selecting then the order does not matter, and if you are arranging the order most likely matters. One example is 10 people running a marathon, how many ways can 3 winners be decided? => for this you only need to select 3 winners so just 10C3, the order does not matter. Now, if it asked, how many different possible ways of winning Gold, Silver, Bronze? => you first need to select 3 winners 10C3, and then arrange them winning G, S, and B 3! => 10C3*3!

ManishaPrepMinds

Tutor

Joined: 14 Apr 2020

Last visit: 24 Apr 2026

Posts: 140

Own Kudos:

155

Given Kudos: 13

Location: India

Expert

Expert reply

Posts: 140

Kudos: 155

19 Aug 2025, 02:56

Kudos

Bookmarks

Rule of thumb

If yes, you are arranging (order matters) → use permutations/arrangements/slot method.
If no, you are selecting (order doesn’t matter) → use combinations.
How to tell quickly
Positions or roles:
Words like arrange, order, sequence, rank, lineup, schedule, seat, assign titles (President/VP), 1st–2nd–3rd, Monday–Tuesday slots → arranging.
Unlabeled groups:
Words like choose, select, pick, group, committee, team, set, hand (cards), bundle → selecting.

Swap test (quick check)

Imagine the same members A, B, C.
- If ABC vs. BAC is a different scenario (e.g., different seats or ranks), it’s arranging
- If ABC vs. BAC is the same scenario (e.g., the same committee), it’s selecting.

Clarity comes with practice. Start by applying these rules to a few easy and medium-level problems, then move on to harder questions to build speed and confidence

Pablopikachu

YashaswaniL

Joined: 08 Oct 2025

Last visit: 29 Apr 2026

Posts: 7

Own Kudos:

Given Kudos: 1

Location: India

Schools: LBS '26

GPA: 8.03

Products:

Schools: LBS '26

Posts: 7

Kudos: 2

17 Apr 2026, 12:37

Kudos

Bookmarks

For the alternate part why can we not solve using 1 - 3C1*3C1*3C1*6C1/9C4 where we chose one from all the three types of tea is denoted by 3C1 and 6C1 implies that we can choose any from rest of the remaining 6 cups. Can anyone please help me with this?

Bunuel

Official Solution:

In a blind taste competition, there are 3 types of tea: Type A, Type B, and Type C. Each type is presented in 3 cups, for a total of 9 cups. If a contestant tastes 4 different cups of tea at random, what is the probability that the contestant does not taste all 3 types of tea?

A. $\frac{1}{12}$
B. $\frac{5}{14}$
C. $\frac{4}{9}$
D. $\frac{1}{2}$
E. $\frac{2}{3}$

To ensure that the contestant does not taste all three types of tea, he must taste only two types of tea. Notice that it's impossible for him to taste just one type of tea, since he will be tasting four cups and in any case, will have to try more than one type of tea.

The number of ways to taste only two types of tea is:

$\frac{C^2_3*C^4_6}{C^4_9}=\frac{5}{14}$:

$C^2_3$ is the number of ways to choose which 2 types of tea out of 3 will be tasted;

$C^4_6$ is the number of ways to choose 4 cups out of 6 cups from these 2 types of tea (2 types = 6 cups);

$C^4_9$ is the total number of ways to choose 4 cups out of 9.

Another way:

Calculate the probability of opposite event and subtract this value from 1.

The opposite event will be for the contestant to taste ALL 3 samples, so the contestant must choose 2 cups from one type of tea, and 1 cup from each of the other two types (2-1-1).

$C^1_3$ is the number of ways to choose the type of tea which will provide with 2 cups;

$C^2_3$ is the number of ways to choose these 2 cups from the chosen type;

$C^1_3$ is the number of ways to choose 1 cup out of 3 from the second type;

$C^1_3$ is the number of ways to choose 1 cup out of 3 from the third type;

$C^4_9$ is the total number of ways to choose 4 cups out of 9.

$P=1-\frac{C^1_3*C^2_3*C^1_3*C^1_3}{C^4_9}=1-\frac{9}{14}=\frac{5}{14}$.

Answer: B