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A certain club has 20 members. What is the ratio of the memb [#permalink]
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25 Mar 2009, 23:04
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This topic is locked. If you want to discuss this question please repost it in the respective forum. A certain club has 20 members. What is the ratio of the member of 5member committees that can be formed from the members of the club to the number of 4member 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]
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25 Mar 2009, 23:17
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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 5member committees that can be formed from the members of the club to the number of 4member 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]
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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: Permutaion + Combination [#permalink]
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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: Permutaion + Combination [#permalink]
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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: Permutaion + Combination [#permalink]
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24 Apr 2010, 10:11
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]
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19 Dec 2010, 15:46
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]
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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: Permutaion + Combination [#permalink]
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31 Aug 2013, 06:38
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]
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03 May 2015, 21:29
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.
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Re: A certain club has 20 members. What is the ratio of the memb [#permalink]
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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 memb [#permalink]
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27 Jul 2015, 06:08
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 memb [#permalink]
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27 Jul 2015, 06:13
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 5member committees that can be formed from the members of the club to the number of 4member 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 memb [#permalink]
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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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A certain club has 20 members. What is the ratio of the memb [#permalink]
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Re: A certain club has 20 members. What is the ratio of the memb [#permalink]
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27 Jul 2015, 06:29
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 memb [#permalink]
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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 memb [#permalink]
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10 Feb 2016, 01:41
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 memb [#permalink]
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06 Jun 2016, 08:41
5 member committee = 20C5 4 member committee = 20C4 Ratio = 20C5 / 20C4 = (20!/15!*5!) * (16!*4!/20!) = 16: 5 Correct option: C
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