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# A five-member committee is to be formed from a group of five

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A five-member committee is to be formed from a group of five [#permalink]

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11 Nov 2007, 14:37
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 18 Jul 2013, 06:11, edited 2 times in total.
Edited the question and added the OA

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Re: A five-member committee is to be formed from a group of five [#permalink]

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20 Jul 2013, 00:07
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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}.

Hope it's clear.
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Re: A five-member committee is to be formed from a group of five [#permalink]

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22 Jul 2015, 20:27
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shyambalaji

The problem is that when you count the number of ways the two spots can be filled with the military personnel you compute it as 9C2 which is 36, however these 36 pairs include all possible combinations from the set of 9 personnel, meaning if the set of military personal is m1, m2, m3, etc, then the pairs will be {m3, m7}, {m1, m8}, etc, however when you select the fifth member you will be repeating some of the same personnel. In other words, you are over counting. For example, the way you are counting will lead to sets such as {c1, c5, m3, m7, m3}, which is not acceptable.

It is best to follow the approach described in the answer by Fasttrack.

Dabral

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Re: PS: Perm & Combination Problem [#permalink]

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18 Jul 2013, 06:12
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Maxirosario2012 wrote:
srivas 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?

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: $$C^3_5*C^2_9=360$$;

Total: 840+360=1,200.

Hope it's clear.
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Re: A five-member committee is to be formed from a group of five [#permalink]

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22 Jul 2013, 11:12
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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..!!!

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Re: A five-member committee is to be formed from a group of five [#permalink]

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22 Jul 2015, 20:55
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shyambalaji wrote:
I tried to follow this approach
First 2 Military Spots = 5C2 = 10
Next 2 Civilian Spots = 9C2 = 36
Last spot can be filled by any of the 10 folks remaining 10C1 = 10
Multiplying all I get 3600. What am I doing wrong here ?

Hi,
I'll try to explain here...
for ease,
let the mil officers be m1,m2,m3,m4,m5..
and civ be c1,c2,c3,c4,c5,c6,c7,c8,c9......

lets take a scenario where you pick up two each from mil and civ say m1,m2,c1,c2 and from remaining 10 say m3
now in your calculations the same set of people will be taken three times...
1)m1,m2,c1,c2 and from remaining 10, m3
2)m1,m3,c1,c2 and from remaining 10, m2
3)m2,m3,c1,c2 and from remaining 10, m1

so you see you have taken same set three times in your calculations

Hope it helped
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11 Nov 2007, 19:01
jimmyjamesdonkey wrote:

Two possibilities:
1. 3 officers and 2 civilians: 5 C 3 * 9 C 2
2. 2 officers and 3 civilians: 5 C 2 * 9 C 3

Total possibilities = 5 C 3 * 9 C 2 + 5 C 2 * 9 C 3

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Re: PS: Perm & Combination Problem [#permalink]

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27 Aug 2008, 17:46
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?

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Re: PS: Perm & Combination Problem [#permalink]

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29 Mar 2009, 01:57
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

Total committees: 180 + 420 = 600

this is analogous to: combination-55369.html

Walker, Suresh, is that correct?

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Re: PS: Perm & Combination Problem [#permalink]

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29 Mar 2009, 11:54
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It's easy to read the question wrong (I think it's purposefully built this way), but once you realize there are 2 possibilities, the rest is easy:

2 officers and 3 civilians
or
3 officers and 2 civilians

9!/6!3!*5!/3!2! + 5!/2!3!*9!/7!2!=1200

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Re: PS: Perm & Combination Problem [#permalink]

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28 Sep 2009, 10:38
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

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Re: PS: Perm & Combination Problem [#permalink]

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18 Jul 2013, 05:53
srivas 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?

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?
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Re: A five-member committee is to be formed from a group of five [#permalink]

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19 Jul 2013, 12:52
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)

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Re: A five-member committee is to be formed from a group of five [#permalink]

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19 Jul 2013, 21:28
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 ?

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Re: A five-member committee is to be formed from a group of five [#permalink]

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20 Jul 2013, 00:11
So basically, we have to divide it by 3, as every case is repeating three times, thanks bunuel.........

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Re: A five-member committee is to be formed from a group of five [#permalink]

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21 Jul 2013, 18:43
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)]

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Re: A five-member committee is to be formed from a group of five [#permalink]

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22 Jul 2015, 17:00
I tried to follow this approach
First 2 Military Spots = 5C2 = 10
Next 2 Civilian Spots = 9C2 = 36
Last spot can be filled by any of the 10 folks remaining 10C1 = 10
Multiplying all I get 3600. What am I doing wrong here ?

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Re: A five-member committee is to be formed from a group of five [#permalink]

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23 Jul 2015, 21:43
{5-military, 9-civilians} - 5 member {2-military, 2-civilians}

The possible cases are:
3-military & 2-civilians - $$5C3 * 9C2 = \frac{5*4*3}{1*2*3} * \frac{9*8}{1*2} = 10 * 36 = 360$$.

2-military & 3-civilians - $$5C2 * 9C3 = \frac{5*4}{1*2} * \frac{9*8*7}{1*2*3} = 10 * 84 = 840$$.

Total number of possibilities = 360+840 = 1200. Ans (B).
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Re: A five-member committee is to be formed from a group of five [#permalink]

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12 Nov 2017, 08:32
Quote:

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

We have two scenarios:

1) 2 civilians and 3 officers; 2) 3 civilians and 2 officers

2 civilians can be selected in 9C2 = 9!/[2!(9-2)!] = 9!/(2!7!) = (9 x 8)/2! = 36 ways

3 officers can be selected in 5C3 = 5!/[3!(5-3)!] = 5!/(3!2!) = (5 x 4 x 3)/3! = 5 x 4 x 3)/(3 x 2 x 1) = 10 ways

So, the total number of possible selections for scenario 1 is 36 x 10 = 360 ways.

Let’s move to scenario 2:

3 civilians can be selected in 9C3 = 9!/[3!(9-3)!] = 9!/(3!6!) = (9 x 8 x 7)/3! = (9 x 8 x 7)/(3 x 2 x 1) = 84 ways

2 officers can be selected in 5C2 = 5!/[2!(5-2)!] = 5!/(2!3!) = (5 x 4)/2! = 10 ways

So, the total number of possible selections for scenario 2 is 84 x 10 = 840 ways.

Thus, the number of ways to make the selections is 840 + 360 = 1,200 ways.

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# A five-member committee is to be formed from a group of five

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