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A five-member committee is to be formed from a group of five [#permalink]
11 Nov 2007, 13:37

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This post was BOOKMARKED

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Difficulty:

65% (hard)

Question Stats:

56% (02:42) correct
44% (01:17) wrong based on 152 sessions

A five-member committee is to be formed from a group of five military officers and nine civilians. If the committee must include at least two officers and two civilians, in how many different ways can the committee be chosen?

Re: PS: Perm & Combination Problem [#permalink]
27 Aug 2008, 16:46

x2suresh wrote:

jimmyjamesdonkey wrote:

A five-member committee is to be formed from a group of five military officers and nine civilians. If the committee must include at least two officers and two civilians, in how many different ways can the committee be chosen?

= 5 C 3 * 9 C 2 + 5 C 2 * 9 C 3 =1200

Question: The reason we donot divide the result by 14 C 5 as we are asked to find "how many different ways can the committee be chose" and not how many "different committees can be chosen?" Correct?

Re: PS: Perm & Combination Problem [#permalink]
29 Mar 2009, 00:57

manOnFire wrote:

x2suresh wrote:

jimmyjamesdonkey wrote:

A five-member committee is to be formed from a group of five military officers and nine civilians. If the committee must include at least two officers and two civilians, in how many different ways can the committee be chosen?

= 5 C 3 * 9 C 2 + 5 C 2 * 9 C 3 =1200

Question: The reason we donot divide the result by 14 C 5 as we are asked to find "how many different ways can the committee be chose" and not how many "different committees can be chosen?" Correct?

No, I think if you ask "how many different committees can be chosen" you have to divide by the permutation of the chosen groups.

thus: Number of different committees:

Different committees of 3 officers / 2 civilians: (5 C 3 * 9 C 2) / 2! = (10 x 36) / 2 = 180

Different committes of 2 officers / 3 civilians: (5 C 2 * 9 C 3) / 2! = (10 x 84) / 2 = 420

A five-member committee is to be formed from a group of five military officers and nine civilians. If the committee must include at least two officers and two civilians, in how many different ways can the committee be chosen?

A five-member committee is to be formed from a group of five military officers and nine civilians. If the committee must include at least two officers and two civilians, in how many different ways can the committee be chosen?

Soln: 5C2 * 9C3 + 5C3 * 9C2

I reached the same solution. Is it right? Or the correct answer is the one of DaveGG who suggests that we need to divide by divide by the permutation of the chosen groups --> 2! ?

The correct answer is 600 or 1200? _________________

Encourage cooperation! If this post was very useful, kudos are welcome "It is our attitude at the beginning of a difficult task which, more than anything else, will affect It's successful outcome" William James

A five-member committee is to be formed from a group of five military officers and nine civilians. If the committee must include at least two officers and two civilians, in how many different ways can the committee be chosen?

Soln: 5C2 * 9C3 + 5C3 * 9C2

I reached the same solution. Is it right? Or the correct answer is the one of DaveGG who suggests that we need to divide by divide by the permutation of the chosen groups --> 2! ?

The correct answer is 600 or 1200?

A five-member committee is to be formed from a group of five military officers and nine civilians. If the committee must include at least two officers and two civilians, in how many different ways can the committee be chosen? A. 119 B. 1,200 C. 3,240 D. 3,600 E. 14,400

To meet the conditions we can have only 2 cases:

2 officers and 3 civilians: \(C^2_5*C^3_9=840\); 3 officers and 2 civilians: \(C^3_5*C^2_9=360\);

Re: A five-member committee is to be formed from a group of five [#permalink]
19 Jul 2013, 23:07

1

This post received KUDOS

Expert's post

abhinawster wrote:

sambam wrote:

5 officers in total. We need to choose at least 2...5C2 9 civilians in total. We need to choose at least 2...9C2

Since the 5th person could be either a civilian or an officer we have 10 people left and we need to choose 1...10C1

(10) x (36) x (10) = 3600 (D)

I used the same method and cudnt figure out where I went wrong, can someone comment on this ?

This number has duplications.

Let's consider for example 5 officers: {A, B, C, D, E}. When you choose 2 of them (with \(C^2_5\)) you can get for example the group {A, B}. Next when you choose one from 10 people then you can get one more officer, for example C, so you'll have in the group 3 officers {A, B, C}. Now, if you choose the group {A, C}, with \(C^2_5\) and then choose B from 10 people then you'll basically get the same 3-officer group: {A, B, C}.

Re: A five-member committee is to be formed from a group of five [#permalink]
21 Jul 2013, 17:43

Bunuel wrote:

This number has duplications.

Let's consider for example 5 officers: {A, B, C, D, E}. When you choose 2 of them (with \(C^2_5\)) you can get for example the group {A, B}. Next when you choose one from 10 people then you can get one more officer, for example C, so you'll have in the group 3 officers {A, B, C}. Now, if you choose the group {A, C}, with \(C^2_5\) and then choose B from 10 people then you'll basically get the same 3-officer group: {A, B, C}.

Hope it's clear.

Bunuel,

I made the same duplication mistake.

(5C2)(9C2)(10C1)=(I picked 2 officers from the 5)(Pick 2 civilians from the 9)(and 1 remainder from the remaining 10).

I am still trying to figure out why they are is a duplication. How/Whats the best way to identify your duplicating your answer? Is there a tell? A quick way to identify you have duplicated your number?

For example look at this question where we pick a person/object from the remaining "nongroup"

Given that there are 5 married couples. If we select only 3 people out of the 10, what is the probability that none of them are married to each other?

One solution: 1-[(5C1)(8C1)/(10C3)]=1-[(we pick one couple)(of that couple we picked we pick 2 people)(for the lat spot we pick someone remaining in the group regardless oh which couple they are in)/(#total number of ways to pick 3 people)]

Re: A five-member committee is to be formed from a group of five [#permalink]
22 Jul 2013, 10:12

1

This post received KUDOS

using the anagram method mentioned in the Number properties guide of MGMAT . .. this problem is a cakewalk.. no need to remember Combination or Permutation formulae..!!!

gmatclubot

Re: A five-member committee is to be formed from a group of five
[#permalink]
22 Jul 2013, 10:12

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