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# A student council is to be chosen from a class of 12 students consisti

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Math Expert
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A student council is to be chosen from a class of 12 students consisti  [#permalink]

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03 Jan 2019, 11:42
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55% (hard)

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58% (01:45) correct 42% (01:31) wrong based on 69 sessions

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A student council is to be chosen from a class of 12 students consisting of a president, a vice president, and 3 committee members. How many such councils are possible

(A) $$\frac{12!}{7!5!}$$

(B) $$\frac{12!}{7!3!}$$

(C) $$\frac{12!}{3!5!}$$

(D) $$\frac{12!}{7!}$$

(E) $$12!$$

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Re: A student council is to be chosen from a class of 12 students consisti  [#permalink]

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03 Jan 2019, 11:47
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A student council is to be chosen from a class of 12 students consisting of a president, a vice president, and 3 committee members. How many such councils are possible

Choosing 5 members from 12 = 12C5 = 12!/ (7! 5!)
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Re: A student council is to be chosen from a class of 12 students consisti  [#permalink]

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03 Jan 2019, 12:30
UB001 wrote:
A student council is to be chosen from a class of 12 students consisting of a president, a vice president, and 3 committee members. How many such councils are possible

Choosing 5 members from 12 = 12C5 = 12!/ (7! 5!)

Wouldn't this need to be a permutation because of the different positions? That would be 12!/7!, but adjust for 3 positions being the same 12!/(7!3!). I'm not certain, but that is my initial thought.

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A student council is to be chosen from a class of 12 students consisti  [#permalink]

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03 Jan 2019, 20:37
1
kanakdaga wrote:
nmccull wrote:
UB001 wrote:
A student council is to be chosen from a class of 12 students consisting of a president, a vice president, and 3 committee members. How many such councils are possible

Choosing 5 members from 12 = 12C5 = 12!/ (7! 5!)

Wouldn't this need to be a permutation because of the different positions? That would be 12!/7!, but adjust for 3 positions being the same 12!/(7!3!). I'm not certain, but that is my initial thought.

Posted from my mobile device

I think so too.
a permutation problem is one that is dependent on arrangement. That means : (a,b) is not equal to (b,a).
A combinations problem is one where (a,b) = (b,a).
Here, we are choosing 5 people out of 12.
But a person chosen for president will be a separate combination than to the person chosen as vice president. Thus, arrangement matters.
so its a permutations problem. thus 12P5
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A student council is to be chosen from a class of 12 students consisti  [#permalink]

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03 Jan 2019, 20:51
"I think so too.
a permutation problem is one that is dependent on arrangement. That means : (a,b) is not equal to (b,a).
A combinations problem is one where (a,b) = (b,a).
Here, we are choosing 5 people out of 12.
But a person chosen for president will be a separate combination than to the person chosen as vice president. Thus, arrangement matters.
so its a permutations problem. thus 12P5"

Yes I am pretty confident it is a permutation, but does it need to be adjusted for the fact that there are only 3 unique positions? 12p5 implies 5 unique positions for arrangement, but there are only 3 in this case (president, vice president, and council member).

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Re: A student council is to be chosen from a class of 12 students consisti  [#permalink]

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03 Jan 2019, 21:13
nmccull wrote:
"I think so too.
a permutation problem is one that is dependent on arrangement. That means : (a,b) is not equal to (b,a).
A combinations problem is one where (a,b) = (b,a).
Here, we are choosing 5 people out of 12.
But a person chosen for president will be a separate combination than to the person chosen as vice president. Thus, arrangement matters.
so its a permutations problem. thus 12P5"

Yes I am pretty confident it is a permutation, but does it need to be adjusted for the fact that there are only 3 unique positions? 12p5 implies 5 unique positions for arrangement, but there are only 3 in this case (president, vice president, and council member).

Posted from my mobile device

You mean that choosing the council members would be a combination while for the three positions it is a permutation ?
lets consider people : a , b, c ,d, e

a as president = 1 way

a as vice president = 1way

a,b,c as council members = 1 way
but a,b,c = b, c, a = c, a, b

thus 3 positions as permutation ?
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Re: A student council is to be chosen from a class of 12 students consisti  [#permalink]

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03 Jan 2019, 21:23
kanakdaga wrote:
nmccull wrote:
"I think so too.
a permutation problem is one that is dependent on arrangement. That means : (a,b) is not equal to (b,a).
A combinations problem is one where (a,b) = (b,a).
Here, we are choosing 5 people out of 12.
But a person chosen for president will be a separate combination than to the person chosen as vice president. Thus, arrangement matters.
so its a permutations problem. thus 12P5"

Yes I am pretty confident it is a permutation, but does it need to be adjusted for the fact that there are only 3 unique positions? 12p5 implies 5 unique positions for arrangement, but there are only 3 in this case (president, vice president, and council member).

Posted from my mobile device

You mean that choosing the council members would be a combination while for the three positions it is a permutation ?
lets consider people : a , b, c ,d, e

a as president = 1 way

a as vice president = 1way

a,b,c as council members = 1 way
but a,b,c = b, c, a = c, a, b

thus 3 positions as permutation ?

Yes, that is how I interpreted the scenario.
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A student council is to be chosen from a class of 12 students consisti  [#permalink]

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03 Jan 2019, 21:51
nmccull wrote:
kanakdaga wrote:
nmccull wrote:
"I think so too.
a permutation problem is one that is dependent on arrangement. That means : (a,b) is not equal to (b,a).
A combinations problem is one where (a,b) = (b,a).
Here, we are choosing 5 people out of 12.
But a person chosen for president will be a separate combination than to the person chosen as vice president. Thus, arrangement matters.
so its a permutations problem. thus 12P5"

Yes I am pretty confident it is a permutation, but does it need to be adjusted for the fact that there are only 3 unique positions? 12p5 implies 5 unique positions for arrangement, but there are only 3 in this case (president, vice president, and council member).

Posted from my mobile device

You mean that choosing the council members would be a combination while for the three positions it is a permutation ?
lets consider people : a , b, c ,d, e

a as president = 1 way

a as vice president = 1way

a,b,c as council members = 1 way
but a,b,c = b, c, a = c, a, b

thus 3 positions as permutation ?

Yes, that is how I interpreted the scenario.

check this question out. https://gmatclub.com/forum/a-club-has-1 ... fl=similar
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A student council is to be chosen from a class of 12 students consisti  [#permalink]

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03 Jan 2019, 22:01
Bunuel wrote:
A student council is to be chosen from a class of 12 students consisting of a president, a vice president, and 3 committee members. How many such councils are possible

(A) $$\frac{12!}{7!5!}$$

(B) $$\frac{12!}{7!3!}$$

(C) $$\frac{12!}{3!5!}$$

(D) $$\frac{12!}{7!}$$

(E) $$12!$$

A president can be chosen in 12 ways. Vice president in 11 ways. Remaining 3 committee members can be selected in $$10_C_3$$ ways = $$\frac{10!}{7!3!}$$

So total possible ways are $$\frac{12!}{7!3!}$$

OPTION: B
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Re: A student council is to be chosen from a class of 12 students consisti  [#permalink]

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11 Jan 2019, 08:52
chetan2u or VeritasKarishma could you please explain how is it a Permutation problem? & the solution
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Re: A student council is to be chosen from a class of 12 students consisti  [#permalink]

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11 Jan 2019, 09:35
2
Manat wrote:
chetan2u or VeritasKarishma could you please explain how is it a Permutation problem? & the solution

Hi ..

It is a combination of permutation and combination..

Out of 12 you have to choose a president, vice president and 3 committee members..
So choosing president and vice president is permutation and selection of 3 committee members is combination.

Go step by step..
1) president - any of the 12, so 12 ways
2) vice president - any of the remaining 11, so 11 ways
3) 3 committee members - 3 out of 10, so 10C3..

Total ways = $$12*11*10C3=12*11*\frac{10!}{7!3!}=\frac{12!}{7!3!}$$

B
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Re: A student council is to be chosen from a class of 12 students consisti   [#permalink] 11 Jan 2019, 09:35
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