|
Author |
Message |
|
TAGS:
|
|
|
Manager
Joined: 03 Jun 2008
Posts: 136
Schools: ISB, Tuck, Michigan (Ross), Darden, MBS
Followers: 1
Kudos [?]:
4
[0], given: 0
|
A college admissions committee will grant a certain number [#permalink]
 15 Aug 2008, 03:03
Question Stats:
20% (01:27) correct
79% (00:45) wrong based on 2 sessions
A college admissions committee will grant a certain number of $10,000 scholarships, $5,000 scholarships, and $1,000 scholarships. If no student can receive more than one scholarship, how many different ways can the committee dole out the scholarships among the pool of 10 applicants? (1) In total, six scholarships will be granted. (2) An equal number of scholarships will be granted at each scholarship level
_________________
-----------------------------------------------------------
'It's not the ride, it's the rider'
|
|
|
|
|
|
|
Director
Joined: 27 May 2008
Posts: 554
Followers: 4
Kudos [?]:
107
[0], given: 0
|
Re: Probability Tough One [#permalink]
15 Aug 2008, 03:49
GMBA85 wrote: A college admissions committee will grant a certain number of $10,000 scholarships, $5,000 scholarships, and $1,000 scholarships. If no student can receive more than one scholarship, how many different ways can the committee dole out the scholarships among the pool of 10 applicants?
(1) In total, six scholarships will be granted.
(2) An equal number of scholarships will be granted at each scholarship level statement 1 : we have total number of scholarships, but we dont have breakup of each type of scholarships. not suff statement 2 : we dont have total number of scholarships, not suff combine : we have 2 scholarships of each type... Suff C
|
|
|
|
|
|
Senior Manager
Joined: 06 Apr 2008
Posts: 452
Followers: 1
Kudos [?]:
25
[0], given: 1
|
Re: Probability Tough One [#permalink]
15 Aug 2008, 04:40
durgesh79 wrote: GMBA85 wrote: A college admissions committee will grant a certain number of $10,000 scholarships, $5,000 scholarships, and $1,000 scholarships. If no student can receive more than one scholarship, how many different ways can the committee dole out the scholarships among the pool of 10 applicants?
(1) In total, six scholarships will be granted.
(2) An equal number of scholarships will be granted at each scholarship level statement 1 : we have total number of scholarships, but we dont have breakup of each type of scholarships. not suff statement 2 : we dont have total number of scholarships, not suff combine : we have 2 scholarships of each type... Suff C Do we really need to know distribution of scholarship? It can be 1,2,3 OR 1,3,2 OR 2,2,2 etc. Statement 1) should be sufficient
|
|
|
|
|
|
SVP
Joined: 30 Apr 2008
Posts: 1900
Location: Oklahoma City
Schools: Hard Knocks
Followers: 25
Kudos [?]:
341
[0], given: 32
|
Re: Probability Tough One [#permalink]
15 Aug 2008, 07:20
nmohindru , We do need to know the number of each scholarship. If you have a variety, there are more options. For example: 3 @ $1k, 2 @ 5k and 1 @ 10k. This could be C_10^3 + C_10_2 + C_10_1 because you need to know how many ways each of the scholarships could be given. Compare this to: All $5k scholarships would be C_10^6. The different scholarships acts somewhat as a differentiator by order. I'm not sure what the final answer is for how many ways, but I'm pretty sure the answer is C to the DS question. GMBA85 wrote: A college admissions committee will grant a certain number of $10,000 scholarships, $5,000 scholarships, and $1,000 scholarships. If no student can receive more than one scholarship, how many different ways can the committee dole out the scholarships among the pool of 10 applicants?
(1) In total, six scholarships will be granted.
(2) An equal number of scholarships will be granted at each scholarship level
_________________
------------------------------------
J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.
Find out what's new at GMAT Club - latest features and updates
|
|
|
|
|
|
Senior Manager
Joined: 31 Jul 2008
Posts: 309
Followers: 1
Kudos [?]:
12
[0], given: 0
|
Re: Probability Tough One [#permalink]
15 Aug 2008, 15:06
A
total no. of ways for choosing 6 people = 10C6
and the number 6 for the three scholarship can be : 222 -- 1 way
123 --- > 3! ways
so answer can be : (1+3!)*10C6
|
|
|
|
|
|
Manager
Joined: 15 Jul 2008
Posts: 213
Followers: 3
Kudos [?]:
14
[0], given: 0
|
Re: Probability Tough One [#permalink]
15 Aug 2008, 15:31
durgesh79 wrote: GMBA85 wrote: A college admissions committee will grant a certain number of $10,000 scholarships, $5,000 scholarships, and $1,000 scholarships. If no student can receive more than one scholarship, how many different ways can the committee dole out the scholarships among the pool of 10 applicants?
(1) In total, six scholarships will be granted.
(2) An equal number of scholarships will be granted at each scholarship level statement 1 : we have total number of scholarships, but we dont have breakup of each type of scholarships. not suff statement 2 : we dont have total number of scholarships, not suff combine : we have 2 scholarships of each type... Suff C Why is 1) insufficient ? With 1) only, there 10C6 * (3^6) ways of doling out the schols .. choose 6 students in 10C6 ways and each of the 6 can be given one of the 3 types of schol.
|
|
|
|
|
|
Director
Joined: 01 Apr 2008
Posts: 921
Schools: IIM Lucknow (IPMX) - Class of 2014
Followers: 8
Kudos [?]:
124
[0], given: 18
|
Re: Probability Tough One [#permalink]
24 Aug 2009, 00:24
I think this is a very important question concept wise. We always think that selecting r items from a collection of n items can be done in nCr ways.
However, this is true only if r items are of the same type.
If r items are different within themselves, then we get more options since we have to multiply the possibilities of each of these subtypes:)
|
|
|
|
|
|
Manager
Joined: 14 Aug 2009
Posts: 134
Followers: 2
Kudos [?]:
81
[0], given: 13
|
Re: Probability Tough One [#permalink]
24 Aug 2009, 17:31
should be A for 1), the different ways are 10P6, doesn't matter what scholarships are. for 2), doesn't mean anything Answer is A.
_________________
Kudos me if my reply helps!
|
|
|
|
|
|
Director
Joined: 01 Apr 2008
Posts: 921
Schools: IIM Lucknow (IPMX) - Class of 2014
Followers: 8
Kudos [?]:
124
[0], given: 18
|
Re: Probability Tough One [#permalink]
25 Aug 2009, 00:19
flyingbunny wrote: should be A
for 1), the different ways are 10P6, doesn't matter what scholarships are.
for 2), doesn't mean anything
Answer is A. Hi Flyingbunny, That was my take too before I got into the details. stmt1 is not sufficient because 10P6 will work only if the 6 items under consideration are of the SAME type. If we have different types (or varieties) then we will have more number of options for selection:) eg. If there are 5 shirts of same color and you have to select 5 then you don't have many options but to go for the same color. If there are 5 shirts of different colors and you have to select 5 then you have got more choices. Try putting some numbers for different scholarship types and check the difference:)
|
|
|
|
|
|
Manager
Joined: 10 Aug 2009
Posts: 139
Followers: 3
Kudos [?]:
46
[1] , given: 10
|
Re: Probability Tough One [#permalink]
25 Aug 2009, 11:43
1
This post received
KUDOS
Answer is C
X_1\times10000+x_2\times5000+x_3\times1000=Total Money Pool Value
1) Stmt 1: X_1+x_2+x_3=6. Howerver, we do not know how many scholarships of each type were distributed. We can't assume any (x_1,X_2,X_3) such that X_1+x_2+x_3=6, since the equation X_1\times10000+x_2\times5000+x_3\times1000=Total Money Pool Value might be violated.
2) Statement 2 : X_1=x_2=x_3=a not sufficient obviously...we do not know the total number of distributed scholarships and we do not know the total value of the money pool...hence we can't figure out the values...note that if we had the total value of money pool we could solve the problem.
Combined
x_1+x_2+x_3=6,
x_1=x_2=x_3
From here x_1=x_2=x_3=2
Last edited by LenaA on 25 Aug 2009, 22:07, edited 1 time in total.
|
|
|
|
|
|
Manager
Joined: 22 Jul 2009
Posts: 200
Followers: 2
Kudos [?]:
141
[0], given: 18
|
Re: Probability Tough One [#permalink]
16 Oct 2009, 13:42
This is a Manhattan GMAT CAT question. OA is C. Although this is a DS question, I would like to find the value. Combining statements 1 and 2, we have: 2 x 10k 2 x 5k 2 x 1k These have to be distributed among 10 students. Distribute 2x10k among 10 students = ways of picking 2 students from 10 = 10C2 Distribute 2x5k among 8 students left = ways of picking 2 students from 8 = 8C2 Distribute 1x5k among 6 students left = ways of picking 2 students from 6 = 6C2 => 10C2*8C2*6C2 = 45*28*15 = 18,900Have the feeling I'm doing something wrong.
_________________
Please kudos if my post helps.
|
|
|
|
|
|
Manager
Joined: 05 Jun 2009
Posts: 77
Followers: 1
Kudos [?]:
2
[0], given: 1
|
Re: Probability Tough One [#permalink]
16 Oct 2009, 14:57
i thought it was A, but my logic is just because there are 10 people and each person cant get the same scholorship doesnt fully define the problem. Because we know there are 3 different scholorships but dont know if its 6 of each or 2 2 2 than we cant fully calculate the total prob, thus you need statement 2. and the answer is C.
|
|
|
|
|
|
Senior Manager
Affiliations: PMP
Joined: 13 Oct 2009
Posts: 319
Followers: 2
Kudos [?]:
76
[0], given: 37
|
Re: Probability Tough One [#permalink]
16 Oct 2009, 19:51
Economist wrote: flyingbunny wrote: should be A
for 1), the different ways are 10P6, doesn't matter what scholarships are.
for 2), doesn't mean anything
Answer is A. Hi Flyingbunny, That was my take too before I got into the details. stmt1 is not sufficient because 10P6 will work only if the 6 items under consideration are of the SAME type. If we have different types (or varieties) then we will have more number of options for selection:) eg. If there are 5 shirts of same color and you have to select 5 then you don't have many options but to go for the same color. If there are 5 shirts of different colors and you have to select 5 then you have got more choices. Try putting some numbers for different scholarship types and check the difference:) I fell into the trap of answering A too, good explanation Economist, Thanks
_________________
Thanks, Sri
-------------------------------
keep uppp...ing the tempo...
Press +1 Kudos, if you think my post gave u a tiny tip
|
|
|
|
|
|
Intern
Joined: 06 Oct 2010
Posts: 14
Followers: 0
Kudos [?]:
1
[0], given: 3
|
Re: Probability Tough One [#permalink]
28 Oct 2010, 02:27
powerka,
Even I got the answer C and tried to find the number of ways using the following way:
[1] choose any 6 students= 10C6
[2] Now, these 6 prizes(2+2+2) can be distributed to 6 people in 6!/ (2!*2!*2!)
(divinding by 2! since 2 prizes of each of 3 prizes are same)
Thus number of ways: [1] * [2] = 18,900
However, I am not sure if the number of ways is correct.
Please, Quant moderator throw more light on the answer. Have my GMAT in a few days...
|
|
|
|
|
|
GMAT Club team member
Joined: 02 Sep 2009
Posts: 11610
Followers: 1800
Kudos [?]:
9593
[1] , given: 828
|
Re: Probability Tough One [#permalink]
28 Oct 2010, 02:34
1
This post received
KUDOS
ptm30 wrote: powerka,
Even I got the answer C and tried to find the number of ways using the following way:
[1] choose any 6 students= 10C6
[2] Now, these 6 prizes(2+2+2) can be distributed to 6 people in 6!/ (2!*2!*2!)
(divinding by 2! since 2 prizes of each of 3 prizes are same)
Thus number of ways: [1] * [2] = 18,900
However, I am not sure if the number of ways is correct.
Please, Quant moderator throw more light on the answer. Have my GMAT in a few days... Yes, 18,900 is the correct answer. This question was also discussed at: college-adminssion-92603.html?hilit=college%20admissions#p713030Below is my solution from there: x= number of 10,000 scholarships y= number of 5,000 scholarships z= number of 1,000 scholarships n= total number of scholarships granted = x+y+z = number of students who will receive scholarship (as no student can receive more than one scholarship) # different ways the committee can dole out the scholarships among the pool of 10 applicants is C^n_{10}*\frac{(x+y+z)!}{x!y!z!}=C^n_{10}*\frac{n!}{x!y!z!}. C^n_{10} - choosing n students who will be granted the scholarship; \frac{n!}{x!y!z!} - # of ways we can distribute x, y and z scholarships among n students. So we need to know the values of n, x, y and z to answer the question. (1) n=6. Don't know x, y and z. Not sufficient. (2) x=y=z. Don't have the exact values of x, y and z . Not sufficient. (1)+(2) n=6 and x=y=z. As n=6=x+y+z --> x=y=z=2 --> # of ways C^n_{10}*\frac{n!}{x!y!z!}=C^6_{10}*\frac{6!}{2!2!2!}. Sufficient. And again: C^6_{10} - choosing 6 students who will be granted the scholarship; \frac{6!}{2!2!2!} - distributing xxyyzz (two of each) among 6 students, which is the # of permuatrions of 6 letters xxyyzz. Answer: C. Hope it's clear.
_________________
PLEASE READ AND FOLLOW: 11 Rules for Posting!!!
RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory
COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!!
DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!
What are GMAT Club Tests?
25 extra-hard Quant Tests
Find out what's new at GMAT Club - latest features and updates
|
|
|
|
|
|
Intern
Joined: 06 Oct 2010
Posts: 14
Followers: 0
Kudos [?]:
1
[0], given: 3
|
Re: Probability Tough One [#permalink]
28 Oct 2010, 03:18
Wow Bunuel!
U r quick in replying. Thanks!
No wonder why I am your follower!
Kudos!
|
|
|
|
|
|
Manager
Joined: 20 Apr 2010
Posts: 227
Schools: ISB, HEC, Said
Followers: 4
Kudos [?]:
8
[0], given: 28
|
Re: Probability Tough One [#permalink]
30 Oct 2010, 05:36
Hi Bunuel,
Is the approach suggested by powerka correct?
Thanks
|
|
|
|
|
|
VP
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 1400
Followers: 8
Kudos [?]:
84
[0], given: 10
|
Re: Probability Tough One [#permalink]
05 May 2011, 22:34
a. 2,2,2 or 3,2,1 or 1,2,3 like wise. Hence POE. b can be 1,1,1; 2,2,2; 3,3,3. Total number of scholarships to be given we do not know. a+b gives 2,2,2. Hence C
_________________
Visit -- http://www.sustainable-sphere.com/
Promote Green Business,Sustainable Living and Green Earth !!
|
|
|
|
|
|
|
Re: Probability Tough One
[#permalink]
05 May 2011, 22:34
|
|
|
|
|
|
|
|
|
|
|
close
