It is currently 16 Oct 2017, 17:45

### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Combinations Basic doubt

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Manager
Joined: 04 Oct 2011
Posts: 217

Kudos [?]: 62 [0], given: 44

Location: India
Concentration: Entrepreneurship, International Business
GMAT 1: 440 Q33 V13
GPA: 3
Combinations Basic doubt [#permalink]

### Show Tags

13 Jan 2013, 23:00
1
This post was
BOOKMARKED
I'm weak in Combinations & Probability

When i went through GMAT Math Book Combinations i have few queries in these examples

1. There are three marbles: 1 blue, 1 gray and 1 green. In how many ways is it possible to arrange marbles in a row?
Solution is : $$3!$$

2. There is a basket with 5 different fruits inside. A child is given the basket and he may pick out none, one or more of the fruits. In how many ways can this choice be made ?
Solution is : $$2^5$$

Both are combinations only.
Why this difference? Is this because arrangement and selection ?
_________________

GMAT - Practice, Patience, Persistence
Kudos if u like

Kudos [?]: 62 [0], given: 44

e-GMAT Representative
Joined: 02 Nov 2011
Posts: 2311

Kudos [?]: 9023 [1], given: 335

Re: Combinations Basic doubt [#permalink]

### Show Tags

14 Jan 2013, 01:02
1
KUDOS
Expert's post
shanmugamgsn wrote:
I'm weak in Combinations & Probability

When i went through GMAT Math Book Combinations i have few queries in these examples

1. There are three marbles: 1 blue, 1 gray and 1 green. In how many ways is it possible to arrange marbles in a row?
Solution is : $$3!$$

2. There is a basket with 5 different fruits inside. A child is given the basket and he may pick out none, one or more of the fruits. In how many ways can this choice be made ?
Solution is : $$2^5$$

Both are combinations only.
Why this difference? Is this because arrangement and selection ?

Hi,

Q 1. Arrangement- When order is important, then we apply arrangement. The total possible arrangements are calculated via permutation formula- nPr.
Example- 3 marbles - 1 blue(B), gray(R) and 1 green(G) can be arranged like these ways. - BRG, BGR, GBR, GRB, RGB, RBG. So total 6 ways. Which is also given by 3! or 3P3 = 3.2.1 = 6 ways.

Alternatively, we can also understand this concept by following way.

# of ways to fill first place = 3 ( Any one of the 3 marbles can be picked up),
# of ways to fill second place = 2 ( Any one of the remaining 2 marbles can be picked up),
# of ways to fill third place = 1 ( Last marble will be picked up)

So total # of ways = 3.2.1 = 3!= 6 ways.

Selection - Selection is applicable when we do not lay any emphasis on arrangement. Only selection is important.
Example- In how many ways can one select 2 students out of 3 students, named Joe, David, & Luke?
Answer- The possible ways would be JD or DL or JL i.e. 3 ways only. JD or DJ is one and the same thing, hence we should not look into arrangement in this case.

Q 2. This question is a sum of 5 combinations- Selection of no fruit + Selection of 1 fruit + Selection of 2 fruits + Selection of 3 fruits + Selection of 4 fruits + Selection of 5 fruits
A very important word here is all 5 fruits are different.

a. Selection of first fruit = 2 ( It will be selected or not selected)
b. Selection of second fruit = 2 ( It will be selected or not selected)
c. Selection of third fruit = 2 ( It will be selected or not selected)
d. Selection of fourth fruit = 2 ( It will be selected or not selected)
e. Selection of fifth fruit = 2 ( It will be selected or not selected)

So total # of ways would be = a.b.c.d.e = 2.2.2.2.2 = 2^5

Hope this make sense to you.

-Shalabh
_________________

| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com

Kudos [?]: 9023 [1], given: 335

Re: Combinations Basic doubt   [#permalink] 14 Jan 2013, 01:02
Display posts from previous: Sort by

# Combinations Basic doubt

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.