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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]  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]  25 Mar 2009, 22:17
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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]  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]  24 Apr 2010, 08:47
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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?

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]  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]  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]  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]  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]  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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Re: A certain club has 20 members. What is the ratio of the memb [#permalink]  16 Feb 2015, 20:51
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Re: A certain club has 20 members. What is the ratio of the memb [#permalink]  03 May 2015, 20:29
dont remember the formular, gmat dose not test.

to take 5
20*19*18*17*16/5!

Princeton review book is very exccelent at explanation of combination and counting. no book can compete Princeton book in this section. read it to master combination and counting. this section is only a few pages and can be learn in a few hours.

this explanation in the book is concise, deep and comprehensive.
Re: A certain club has 20 members. What is the ratio of the memb   [#permalink] 03 May 2015, 20:29
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