Last visit was: 16 Jul 2024, 12:32 It is currently 16 Jul 2024, 12:32
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.
Close
Request Expert Reply
Confirm Cancel
SORT BY:
Date
Tags:
Show Tags
Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 94370
Own Kudos [?]: 641398 [7]
Given Kudos: 85332
Send PM
Retired Moderator
Joined: 10 Nov 2018
Posts: 536
Own Kudos [?]: 469 [1]
Given Kudos: 229
Location: India
Concentration: General Management, Strategy
GMAT 1: 590 Q49 V22
WE:Other (Retail)
Send PM
VP
VP
Joined: 14 Jul 2020
Posts: 1115
Own Kudos [?]: 1300 [2]
Given Kudos: 351
Location: India
Send PM
Target Test Prep Representative
Joined: 14 Oct 2015
Status:Founder & CEO
Affiliations: Target Test Prep
Posts: 19148
Own Kudos [?]: 22646 [1]
Given Kudos: 286
Location: United States (CA)
Send PM
Re: A box contains two white balls, three black balls and four red balls. [#permalink]
1
Kudos
Expert Reply
Bunuel wrote:
A box contains two white balls, three black balls and four red balls. In how many ways can three balls be drawn from the box, if at least one black ball is to be included in the draw?

A. 16
B. 32
C. 64
D. 96
E. 128



Solution:

Without any restrictions, the number of ways to choose 3 balls from 9 is 9C3 = (9 x 8 x 7)/(3 x 2) = 3 x 4 x 7 = 84.

Let’s calculate the number of ways we don’t select a black ball. Since there are 6 non-black balls, we can choose any 3 of them in 6C3 = (6 x 5 x 4) / (3 x 2) = 120/6 = 20 ways.

In summary, there are 84 different selections of 3 balls (if there are no restrictions), and 20 of them consist of selecting no black balls. Therefore, there must be 84 - 20 = 64 different selection of 3 balls with at least one black ball.

Answer: C
GMAT Club Legend
GMAT Club Legend
Joined: 03 Oct 2013
Affiliations: CrackVerbal
Posts: 4918
Own Kudos [?]: 7802 [0]
Given Kudos: 220
Location: India
Send PM
Re: A box contains two white balls, three black balls and four red balls. [#permalink]
Top Contributor
A box contains two white balls, three black balls and four red balls. In how many ways can three balls be drawn from the box, if at least one black ball is to be included in the draw?

Total 9 balls are there ( 2W, 3B,4R)

Total no of ways to select 3 balls out of 9 = 9C3 = 9*8*7/3*2*1 = 84

Total no of ways to select 3 balls ,where there is no black ball selected i.e you should select from either white or red balls ( 2W + 4R) = 6C3= 20

Total no of ways to select 3 balls if atleast 1 black ball is included = Total no of ways to select 3 balls out of 9 - Total no of ways to select 3 balls ,where there is no black ball selected = 84 -20 = 64

Option C is the answer.

Thanks,
Clifin J Francis.
Senior Manager
Senior Manager
Joined: 11 Sep 2022
Posts: 497
Own Kudos [?]: 175 [0]
Given Kudos: 2
Location: India
Paras: Bhawsar
GMAT 1: 590 Q47 V24
GMAT 2: 580 Q49 V21
GMAT 3: 700 Q49 V35
GPA: 3.2
WE:Project Management (Other)
Send PM
Re: A box contains two white balls, three black balls and four red balls. [#permalink]
Total no. ways to select atleast 1 black ball = total ways of selection of any 3 balls from 9 balls without any restriction - total ways of selection 3 balls from group of only red and white balls
=9C3-6C3
=84-20
=64
GMAT Club Bot
Re: A box contains two white balls, three black balls and four red balls. [#permalink]
Moderator:
Math Expert
94370 posts