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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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Question Stats:

71% (01:56) correct
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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

Re: Permutaion + Combination [#permalink]
25 Mar 2009, 22:17

1

This post was BOOKMARKED

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

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.

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]

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? _________________

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, ... _________________

A certain club has 20 members. What is the ratio of the [#permalink]
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?

Re: A certain club has 20 members. What is the ratio of the [#permalink]
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?

Re: A certain club has 20 members. What is the ratio of the memb [#permalink]
16 Feb 2015, 20:51

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