For a scholarship, at most n candidates out of 2n + 1 can be : GMAT Problem Solving (PS)
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 20 Jan 2017, 10:29

### 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

# For a scholarship, at most n candidates out of 2n + 1 can be

Author Message
TAGS:

### Hide Tags

CEO
Joined: 15 Aug 2003
Posts: 3460
Followers: 67

Kudos [?]: 862 [0], given: 781

For a scholarship, at most n candidates out of 2n + 1 can be [#permalink]

### Show Tags

17 Sep 2003, 01:52
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 1 sessions

### HideShow timer Statistics

For a scholarship, at most n candidates out of 2n + 1 can be selected. If the number of different ways of selection of at least one candidate is 63, the maximum number of candidates that can be selected for the scholarship is:

1. 3
2. 4
3. 2
4. 5
Intern
Joined: 17 Sep 2003
Posts: 21
Location: GMAT Maze, Chaos.
Followers: 0

Kudos [?]: 12 [0], given: 0

### Show Tags

18 Sep 2003, 06:11
Do u know the answer to the problem ?

I m not able to solve it, the qn is kinda confusing.
Intern
Joined: 28 Aug 2003
Posts: 36
Location: USA
Followers: 0

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

### Show Tags

18 Sep 2003, 16:36
I could do this only by backsolving.

63 = C(2n+1,1) + C(2n+1,2) + .... + C(2n+1,n)

Question stem is asking for n.

A. n = 3 then 2n+1 = 7.

7C1 + 7C2 + 7C3 = 7 + 21 + 35 = 63.

There has to be a better way, I think. Anyone? Vicks? Praet?
CEO
Joined: 15 Aug 2003
Posts: 3460
Followers: 67

Kudos [?]: 862 [0], given: 781

### Show Tags

19 Sep 2003, 01:56
edealfan wrote:
I could do this only by backsolving.

63 = C(2n+1,1) + C(2n+1,2) + .... + C(2n+1,n)

Question stem is asking for n.

A. n = 3 then 2n+1 = 7.

7C1 + 7C2 + 7C3 = 7 + 21 + 35 = 63.

There has to be a better way, I think. Anyone? Vicks? Praet?

Honestly, i cant think of a better way.
Vicks , can you help with this.
Manager
Joined: 23 May 2013
Posts: 127
Followers: 1

Kudos [?]: 55 [0], given: 110

Re: For a scholarship, at most n candidates out of 2n + 1 can be [#permalink]

### Show Tags

26 Feb 2014, 09:18
Praetorian wrote:
For a scholarship, at most n candidates out of 2n + 1 can be selected. If the number of different ways of selection of at least one candidate is 63, the maximum number of candidates that can be selected for the scholarship is:

A. 3
B. 4
C. 2
D. 5

No of ways to select at least one candidate out of x candidates is 2^x - 1
2^x - 1 =63
2^x = 64
hence X= 6

But we are told that 2n+1 = x
hence n = 2.5 or >2. Hence we choose A
_________________

“Confidence comes not from always being right but from not fearing to be wrong.”

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7125
Location: Pune, India
Followers: 2137

Kudos [?]: 13667 [0], given: 222

Re: For a scholarship, at most n candidates out of 2n + 1 can be [#permalink]

### Show Tags

26 Feb 2014, 18:58
ankur1901 wrote:
Praetorian wrote:
For a scholarship, at most n candidates out of 2n + 1 can be selected. If the number of different ways of selection of at least one candidate is 63, the maximum number of candidates that can be selected for the scholarship is:

A. 3
B. 4
C. 2
D. 5

No of ways to select at least one candidate out of x candidates is 2^x - 1
2^x - 1 =63
2^x = 64
hence X= 6

But we are told that 2n+1 = x
hence n = 2.5 or >2. Hence we choose A

This is not correct. Getting a fractional value for n should be the clue since n = 2.5 could mean n = 2 candidates.

Remember that at most n candidates can be selected. The number of ways you can select at least 1 candidate up to n candidates out of the total 2n+1 is given as 63.

$$(2n+1)C0 + (2n+1)C1 + (2n+1)C2 + ... + (2n+1)C(n) + (2n+1)C(n+1) + (2n+1)C(n+2) + ... + (2n+1)C(2n+1) = 2^{2n+1}$$

Note that you are given that $$(2n+1)C1 + (2n+1)C2 + ... + (2n+1)C(n) = 63$$. We also know $$(2n+1)C0 = 1.$$
Also recall that $$(2n+1)C0 + (2n+1)C1 + (2n+1)C2 + ... + (2n+1)C(n) = (2n+1)C(n+1) + (2n+1)C(n+2) + ... + (2n+1)C(2n+1)$$
since nCr = nC(n-r)

$$1 + 63 + 63 + 1 = 2^{2n+1}$$
$$n = 3$$

For non-engineering students, this can be solved by backsolving though that's usually time consuming so I doubt GMAT will test such a concept.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Re: For a scholarship, at most n candidates out of 2n + 1 can be   [#permalink] 26 Feb 2014, 18:58
Similar topics Replies Last post
Similar
Topics:
4 If 3^(2n) = (1/9)^(n+2), what is the value of n? 3 21 Mar 2016, 06:38
. Consider the sets Tn = {n, n + 1, n + 2, n + 3, n + 4}, where n = 1 2 08 Nov 2015, 00:41
7 If 12^(1/2) + 108^(1/2) = N^(1/2), then N = 10 13 Mar 2015, 06:46
7 If n is a positive integer greater than 1, then 2^{n-1} + 2^ 7 06 Sep 2013, 23:46
17 If (n-2)! = (n! + (n-1)!)/99, and n is a positive integer, then n=? 14 21 Nov 2010, 18:21
Display posts from previous: Sort by