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

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milind1979
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A certain club has 20 members. What is the ratio of the number of [#permalink] New post   25 Mar 2009, 23:04
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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?

A. 16 to 1
B. 15 to 1
C. 16 to 5
D. 15 to 6
E. 5 to 4
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C
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Bunuel
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Re: A certain club has 20 members. What is the ratio of the number of [#permalink] New post   24 Apr 2010, 10:05
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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: A certain club has 20 members. What is the ratio of the number of [#permalink] New post   25 Mar 2009, 23:17
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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: A certain club has 20 members. What is the ratio of the number of [#permalink] New post   24 Apr 2010, 09:38
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: A certain club has 20 members. What is the ratio of the number of [#permalink] New post   24 Apr 2010, 09:47
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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: A certain club has 20 members. What is the ratio of the number of [#permalink] New post   29 Aug 2013, 11:11
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: A certain club has 20 members. What is the ratio of the number of [#permalink] New post   31 Aug 2013, 06:38
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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 number of [#permalink] New post   27 Jul 2015, 06:01
Isnt the Question doubtful?

They didnt mentioned that first we have to select 5 members then 4 from the remaining 15?
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Re: A certain club has 20 members. What is the ratio of the number of [#permalink] New post   27 Jul 2015, 06:08
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honchos wrote:
Isnt the Question doubtful?

They didnt mentioned that first we have to select 5 members then 4 from the remaining 15?


That is exactly the reason why you need to treat "selecting 5 out of 20" and "selecting 4 out of 20" as 2 independent events. Per the question, there is no overlap in the 2 cases.

Thus the final answer = 20C5 / 20C4 = 16/5
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Re: A certain club has 20 members. What is the ratio of the number of [#permalink] New post   27 Jul 2015, 06:13
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honchos wrote:
Isnt the Question doubtful?

They didnt mentioned that first we have to select 5 members then 4 from the remaining 15?


The question is absolutely clear and holds no ambiguity in the language

Question : 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?"


The two actions (making the committee of 5 members AND making the committee of 4 embers) are to be treated independently.

No. of Ways to make 5 member committee out of 20 members = 20C5 = 15504

No. of Ways to make 4 member committee out of 20 members = 20C4 = 4845

Required ratio = 15504 / 4845 = 16/5
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Re: A certain club has 20 members. What is the ratio of the number of [#permalink] New post   27 Jul 2015, 06:22
Engr2012 wrote:
honchos wrote:
Isnt the Question doubtful?

They didnt mentioned that first we have to select 5 members then 4 from the remaining 15?


That is exactly the reason why you need to treat "selecting 5 out of 20" and "selecting 4 out of 20" as 2 independent events. Per the question, there is no overlap in the 2 cases.

Thus the final answer = 20C5 / 20C4 = 16/5



Thanks, My doubt is still the same. They didn't mentioned that two events are mutually independent or dependent.
This is a very easy question, but I was spinned that how should I interpret the language - Mutually independent or independent.
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Re: A certain club has 20 members. What is the ratio of the number of [#permalink] New post   27 Jul 2015, 06:29
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honchos wrote:
Engr2012 wrote:
honchos wrote:
Isnt the Question doubtful?

They didnt mentioned that first we have to select 5 members then 4 from the remaining 15?


That is exactly the reason why you need to treat "selecting 5 out of 20" and "selecting 4 out of 20" as 2 independent events. Per the question, there is no overlap in the 2 cases.

Thus the final answer = 20C5 / 20C4 = 16/5



Thanks, My doubt is still the same. They didn't mentioned that two events are mutually independent or dependent.
This is a very easy question, but I was spinned that how should I interpret the language - Mutually independent or independent.


The question did NOT mentioned that "Team of 4 has to be selected out of the REMAINING"

If the word "remaining" has not been used then it is Independent event case.

I hope it helps1
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Re: A certain club has 20 members. What is the ratio of the number of [#permalink] New post   09 Feb 2016, 14:33
Hello guys,

Can I please ask how 20C5 / 20C4 has been solved to get the fraction 16/5?
I know that one way to solve this is by doing the long multiplication thing but Brunnel has calculated it in a simplified manner. Can someone please write down the steps?

Thanks,
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Re: A certain club has 20 members. What is the ratio of the number of [#permalink] New post   10 Feb 2016, 01:41
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HarveyKlaus wrote:
Hello guys,

Can I please ask how 20C5 / 20C4 has been solved to get the fraction 16/5?
I know that one way to solve this is by doing the long multiplication thing but Brunnel has calculated it in a simplified manner. Can someone please write down the steps?

Thanks,


20C5 = 20!/[5! * 15!] = 20*19*18*17*16/5! = 19*3*17*16

20C4 = 20!/[4! * 16!] = 20*19*18*17/4! = 5*19*3*17

20C5 / 20C4 = 19*3*17*16 / 5*19*3*17 = 16/5

I hope this helps!!!
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Re: A certain club has 20 members. What is the ratio of the number of [#permalink] New post   Updated on: 17 May 2021, 08:30
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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


5-member committees
Since the order in which we select the committee members does not matter, we can use COMBINATIONS
We can select 5 people from 20 people in 20C5 ways
20C5 = (20)(19)(18)(17)(16)/(5)(4)(3)(2)(1)

4-member committees
Since the order in which we select the committee members does not matter, we can use COMBINATIONS
We can select 4 people from 20 people in 20C4 ways
20C4 = (20)(19)(18)(17)/(4)(3)(2)(1)

The RATIO = (20)(19)(18)(17)(16)/(5)(4)(3)(2)(1)]/[(20)(19)(18)(17)/(4)(3)(2)(1)]
= (20)(19)(18)(17)(16)/(5)(4)(3)(2)(1)][(4)(3)(2)(1)/(20)(19)(18)(17)]
=(20)(19)(18)(17)(16)(4)(3)(2)(1)/(20)(19)(18)(17)(5)(4)(3)(2)(1)
=(20)(19)(18)(17)(16)(4)(3)(2)(1)/(20)(19)(18)(17)(5)(4)(3)(2)(1)
=16/5

Answer: C

Cheers,
Brent

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Re: A certain club has 20 members. What is the ratio of the number of [#permalink] New post   04 Mar 2019, 20:18
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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


The ratio is the following:

(20C5)/(20C4) = [(20 x 19 x 18 x 17 x 16)/5!]/[(20 x 19 x 18 x 17)/4!]

= [(20 x 19 x 18 x 17 x 16)/5!] x [4!/(20 x 19 x 18 x 17)]

= 16/5

Answer: C
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