At a Pizza Parlor, in addition to cheese there are 10 : PS Archive
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 17 Jan 2017, 00:10

### 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

# At a Pizza Parlor, in addition to cheese there are 10

Author Message
Director
Joined: 14 Jan 2007
Posts: 777
Followers: 2

Kudos [?]: 136 [0], given: 0

At a Pizza Parlor, in addition to cheese there are 10 [#permalink]

### Show Tags

23 Jun 2007, 05:05
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

At a Pizza Parlor, in addition to cheese there are 10 different toppings. If you can order any number of toppings, then how many different toppings are possible.
VP
Joined: 08 Jun 2005
Posts: 1146
Followers: 7

Kudos [?]: 189 [0], given: 0

### Show Tags

23 Jun 2007, 06:12
C(10,0) = 1
C(10,1) = 10
C(10,2) = 45
C(10,3) = 120
C(10,4) = 210
C(10,5) = 252
C(10,6) = 210
C(10,7) = 120
C(10,8) = 45
C(10,9) = 10
C(10,10) = 1

total = 1,024

check = 2^10 = 1,024

Director
Joined: 14 Jan 2007
Posts: 777
Followers: 2

Kudos [?]: 136 [0], given: 0

### Show Tags

23 Jun 2007, 06:17
KillerSquirrel wrote:
C(10,0) = 1
C(10,1) = 10
C(10,2) = 45
C(10,3) = 120
C(10,4) = 210
C(10,5) = 252
C(10,6) = 210
C(10,7) = 120
C(10,8) = 45
C(10,9) = 10
C(10,10) = 1

total = 1,024

check = 2^10 = 1,024

That's correct.
Shortcut for such questions is -
nC0 + nC1 +nC2 +.........nCn = 2^n
Director
Joined: 26 Feb 2006
Posts: 904
Followers: 4

Kudos [?]: 107 [0], given: 0

### Show Tags

23 Jun 2007, 13:29
KillerSquirrel wrote:
C(10,0) = 1
C(10,1) = 10
C(10,2) = 45
C(10,3) = 120
C(10,4) = 210
C(10,5) = 252
C(10,6) = 210
C(10,7) = 120
C(10,8) = 45
C(10,9) = 10
C(10,10) = 1

total = 1,024

check = 2^10 = 1,024

interesting! I never noticed this one.

Is it always true for any n that 2^n?
VP
Joined: 08 Jun 2005
Posts: 1146
Followers: 7

Kudos [?]: 189 [0], given: 0

### Show Tags

23 Jun 2007, 13:54
Himalayan wrote:
KillerSquirrel wrote:
C(10,0) = 1
C(10,1) = 10
C(10,2) = 45
C(10,3) = 120
C(10,4) = 210
C(10,5) = 252
C(10,6) = 210
C(10,7) = 120
C(10,8) = 45
C(10,9) = 10
C(10,10) = 1

total = 1,024

check = 2^10 = 1,024

interesting! I never noticed this one.

Is it always true for any n that 2^n?

Yes ! "In general, the sum of all the combinations of n distinct things is 2^n.

nC0 + nC1 + nC2 + . . . + nCn = 2^n"

Last edited by KillerSquirrel on 24 Jun 2007, 13:27, edited 1 time in total.
Director
Joined: 26 Feb 2006
Posts: 904
Followers: 4

Kudos [?]: 107 [0], given: 0

### Show Tags

23 Jun 2007, 20:00
KillerSquirrel wrote:
Himalayan wrote:
KillerSquirrel wrote:
C(10,0) = 1
C(10,1) = 10
C(10,2) = 45
C(10,3) = 120
C(10,4) = 210
C(10,5) = 252
C(10,6) = 210
C(10,7) = 120
C(10,8) = 45
C(10,9) = 10
C(10,10) = 1

total = 1,024

check = 2^10 = 1,024

interesting! I never noticed this one.

Is it always true for any n that 2^n?

Yes ! "In general, the sum of all the combinations of n distinct things is 2n.

nC0 + nC1 + nC2 + . . . + nCn = 2^n"

grate learning. this concept directly goes to my note.

thanx..
23 Jun 2007, 20:00
Display posts from previous: Sort by