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A certain club has 20 members. What is the ratio of the memb

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A certain club has 20 members. What is the ratio of the memb [#permalink] New post 25 Mar 2009, 22:04
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A certain club has 20 members. What is the ratio of the member of 5-member committees that can be formed from the members of the club to the number of 4-member committees that can be formed from the members of the club?

A. 16 to 1
B. 15 to 1
C. 16 to 5
D. 15 to 6
E. 5 to 4
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Re: Permutaion + Combination [#permalink] New post 25 Mar 2009, 22:17
C.
20C5/20C4 = 16/5
milind1979 wrote:
A certain club has 20 members. What is the ratio of the member of 5-member
committees that can be formed from the members of the club to the number of 4-member
committees that can be formed from the members of the club?
A. 16 to 1
B. 15 to 1
C. 16 to 5
D. 15 to 6
E. 5 to 4
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Re: Permutaion + Combination [#permalink] New post 24 Apr 2010, 08:38
brinng back an old post.

I somehow can't get 20C5/20C4 to be 16/5; I keep on getting 8/5

Also, for this question, why can't we just do the following:
(20*19*18*17*16)/(20*19*18*17) = 16/1?
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Re: Permutaion + Combination [#permalink] New post 24 Apr 2010, 08:47
bakfed wrote:
brinng back an old post.

I somehow can't get 20C5/20C4 to be 16/5; I keep on getting 8/5

Also, for this question, why can't we just do the following:
(20*19*18*17*16)/(20*19*18*17) = 16/1?


20C5 = FACT[20] / FACT[5]*FACT[15]
20C4 = FACT[20] / FACT[4]*FACT[16]

20C5 / 20C4 = 16/5
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Re: Permutaion + Combination [#permalink] New post 24 Apr 2010, 09:05
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bakfed wrote:
brinng back an old post.

I somehow can't get 20C5/20C4 to be 16/5; I keep on getting 8/5

Also, for this question, why can't we just do the following:
(20*19*18*17*16)/(20*19*18*17) = 16/1?


\frac{C^5_{20}}{C^4_{20}}=\frac{20!}{15!*5!}*\frac{16!*4!}{20!}=\frac{16}{5}

Second question:

20*19*18*17*16 does not give you # of 5 member committees out of 20. You need to divide this by 5! to get rid of the repetitions (factorial correction). The same for 20*19*18*17, you should divide this by 4!.

Take another example: how many committees of 2 can be formed out of A, B and C?

AB
AC
BC
Only 3, which is C^2_3=3.

But the way you are doing you'd get 3*2=6. This number has repetitions so we should divide it by 2! --> 6/2!=3.

Hope it's clear.
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Re: Permutaion + Combination [#permalink] New post 24 Apr 2010, 09:11
bakfed wrote:
brinng back an old post.

Also, for this question, why can't we just do the following:
(20*19*18*17*16)/(20*19*18*17) = 16/1?


20*19*18*17*16 is giving all possible ways of selecting 5 people. It also includes the order of a particular selection into account. In this case it does not mater if you select a member first ,second or third as long as it is selected in the group. Therefore, the 20*19*18*17*16 number has repetition of the same group in 5 ways and you have to divide by FACT[5]. Similarly for (20*19*18*17) by FACT[4]
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Re: Permutaion + Combination [#permalink] New post 19 Dec 2010, 14:46
I initially got this wrong, but see that this is just 20!/15!5! *16!4!/20! the 20! cancels out and you have 16!4!/15!5! which leaves 16/5.
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Re: Permutaion + Combination [#permalink] New post 29 Aug 2013, 10:11
Bunuel wrote:
bakfed wrote:
brinng back an old post.

I somehow can't get 20C5/20C4 to be 16/5; I keep on getting 8/5

Also, for this question, why can't we just do the following:
(20*19*18*17*16)/(20*19*18*17) = 16/1?


\frac{C^5_{20}}{C^4_{20}}=\frac{20!}{15!*5!}*\frac{16!*4!}{20!}=\frac{16}{5}

Second question:

20*19*18*17*16 does not give you # of 5 member committees out of 20. You need to divide this by 5! to get rid of the repetitions (factorial correction). The same for 20*19*18*17, you should divide this by 4!.

Take another example: how many committees of 2 can be formed out of A, B and C?

AB
AC
BC
Only 3, which is C^2_3=3.

But the way you are doing you'd get 3*2=6. This number has repetitions so we should divide it by 2! --> 6/2!=3.

Hope it's clear.


when should we use 20*19*18*17*16 way and when the second way?
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Re: Permutaion + Combination [#permalink] New post 31 Aug 2013, 05:38
Expert's post
swati007 wrote:
Bunuel wrote:
bakfed wrote:
brinng back an old post.

I somehow can't get 20C5/20C4 to be 16/5; I keep on getting 8/5

Also, for this question, why can't we just do the following:
(20*19*18*17*16)/(20*19*18*17) = 16/1?


\frac{C^5_{20}}{C^4_{20}}=\frac{20!}{15!*5!}*\frac{16!*4!}{20!}=\frac{16}{5}

Second question:

20*19*18*17*16 does not give you # of 5 member committees out of 20. You need to divide this by 5! to get rid of the repetitions (factorial correction). The same for 20*19*18*17, you should divide this by 4!.

Take another example: how many committees of 2 can be formed out of A, B and C?

AB
AC
BC
Only 3, which is C^2_3=3.

But the way you are doing you'd get 3*2=6. This number has repetitions so we should divide it by 2! --> 6/2!=3.

Hope it's clear.


when should we use 20*19*18*17*16 way and when the second way?


The first case is applicable when the order matters, when, for example, we have member #1, member #2, ...
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A certain club has 20 members. What is the ratio of the [#permalink] New post 30 Nov 2013, 13:41
A certain club has 20 members. What is the ratio of the number of 5-member committees that can be formed from the members of the club to the number of 4 member committees that can be formed from the members of the club?

16 to 1
15 to 1
16 to 5
15 to 6
5 to 4

[Reveal] Spoiler:
PLS explain the OA
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Re: A certain club has 20 members. What is the ratio of the [#permalink] New post 01 Dec 2013, 05:25
Expert's post
MDK wrote:
A certain club has 20 members. What is the ratio of the number of 5-member committees that can be formed from the members of the club to the number of 4 member committees that can be formed from the members of the club?

16 to 1
15 to 1
16 to 5
15 to 6
5 to 4

[Reveal] Spoiler:
PLS explain the OA


Merging similar topics. Please refer tot he solutions above.
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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; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

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. ,11 Mixed Questions, 12 Fresh Meat

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., 11 New DS set.


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Re: A certain club has 20 members. What is the ratio of the   [#permalink] 01 Dec 2013, 05:25
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