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Eight chairs are numbered 1 to 8. Two women and three men wish to occu

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Bunuel
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Eight chairs are numbered 1 to 8. Two women and three men wish to occu [#permalink] New post   18 Feb 2021, 07:30
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Eight chairs are numbered 1 to 8. Two women and three men wish to occupy one chair each. First the women choose the chairs from amongst the chairs marked 1 to 4, and then the men select the chairs from amongst the remaining. What is the number of possible arrangements ?

(A) 6C3 * 4C2
(B) 4P2 * 4P3
(C) 4C2 + 4P3
(D) 4P2*6P3
(E) 4C2


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Eight chairs are numbered 1 to 8. Two women and three men wish to occu [#permalink] New post   Updated on: 28 Feb 2021, 05:41
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Bunuel wrote:
Eight chairs are numbered 1 to 8. Two women and three men wish to occupy one chair each. First the women choose the chairs from amongst the chairs marked 1 to 4, and then the men select the chairs from amongst the remaining. What is the number of possible arrangements ?

(A) 6C3 * 4C2
(B) 4P2 * 4P3
(C) 4C2 + 4P3
(D) 4P3
(E) 4C2



The women may select 2 chairs out of 4 (numbered 1 to 4) in 4C2 ways
They can occupy those two chairs in 2! ways

Now, men have 6 chairs to choose 3 from which they can select in = 6C3 ways
Now the men can occupy the selected chairs in 3! ways so

Answer = 4C2*2! * 6C3*3! = 4P2*6P3

Answer: Option D

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Originally posted by GMATinsight on 18 Feb 2021, 07:40.
Last edited by GMATinsight on 28 Feb 2021, 05:41, edited 4 times in total.
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Re: Eight chairs are numbered 1 to 8. Two women and three men wish to occu [#permalink] New post   28 Feb 2021, 04:58
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Two women can be seated amongst chairs marked 1 to 4 = 4P2
Three men can be seated amongst the remaining 6 chairs = 6P3.
Total number of arrangement = 4P2*6P3


Bunuel wrote:
Eight chairs are numbered 1 to 8. Two women and three men wish to occupy one chair each. First the women choose the chairs from amongst the chairs marked 1 to 4, and then the men select the chairs from amongst the remaining. What is the number of possible arrangements ?

(A) 6C3 * 4C2
(B) 4P2 * 4P3
(C) 4C2 + 4P3
(D) 4P3
(E) 4C2


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Re: Eight chairs are numbered 1 to 8. Two women and three men wish to occu [#permalink] New post   28 Feb 2021, 11:27
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GMATBusters wrote:
Two women can be seated amongst chairs marked 1 to 4 = 4P2
Three men can be seated amongst the remaining 6 chairs = 6P3.
Total number of arrangement = 4P2*6P3


Bunuel wrote:
Eight chairs are numbered 1 to 8. Two women and three men wish to occupy one chair each. First the women choose the chairs from amongst the chairs marked 1 to 4, and then the men select the chairs from amongst the remaining. What is the number of possible arrangements ?

(A) 6C3 * 4C2
(B) 4P2 * 4P3
(C) 4C2 + 4P3
(D) 4P3
(E) 4C2


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Thanks GMATBusters.

Your explanations are always so precise and clear. I wonder how easy things look after your explanation.

Thank you once again

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Re: Eight chairs are numbered 1 to 8. Two women and three men wish to occu [#permalink] New post   19 Feb 2021, 07:15
GMATinsight wrote:
Bunuel wrote:
Eight chairs are numbered 1 to 8. Two women and three men wish to occupy one chair each. First the women choose the chairs from amongst the chairs marked 1 to 4, and then the men select the chairs from amongst the remaining. What is the number of possible arrangements ?

(A) 6C3 * 4C2
(B) 4P2 * 4P3
(C) 4C2 + 4P3
(D) 4P3
(E) 4C2



The women may select 2 chairs out of 4 (numbered 1 to 4) in 4C2 ways
They can occupy those two chairs in 2! ways

Now, men have 6 chairs to choose 3 from which they can select in = 6C3 ways
Now the men can occupy the selected chairs in 3! ways so

Answer = 4C2*2! * 6C3*3! = 4P2*6P3

Answer: Option B




IMO "amongs remaining" means those not marked 1 to 4, so 2 men may select from those 4 not marked chairs.
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Re: Eight chairs are numbered 1 to 8. Two women and three men wish to occu [#permalink] New post   19 Feb 2021, 07:43
GMATinsight wrote:
Bunuel wrote:
Eight chairs are numbered 1 to 8. Two women and three men wish to occupy one chair each. First the women choose the chairs from amongst the chairs marked 1 to 4, and then the men select the chairs from amongst the remaining. What is the number of possible arrangements ?

(A) 6C3 * 4C2
(B) 4P2 * 4P3
(C) 4C2 + 4P3
(D) 4P3
(E) 4C2



The women may select 2 chairs out of 4 (numbered 1 to 4) in 4C2 ways
They can occupy those two chairs in 2! ways

Now, men have 4 chairs to choose 3 from which they can select in = 4C3 ways
Now the men can occupy the selected chairs in 3! ways so

Answer = 4C2*2! * 4C3*3! = 4P2*4P3

Answer: Option B



Could you please explain why the 2 women can select 2 out of 4 chairs in 4C2 ways instead of 4P2? Since the chairs are numbered we can infer that order matters, so first woman can choose between 4 chairs, and second one has 3 choices left. Shouldn't it be then 4*3 = 12 possible selections?

Also if the 3 men can choose between the remaining chairs, then they have 8-2 = 6 choices...

The text of this problem is not super clear
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Re: Eight chairs are numbered 1 to 8. Two women and three men wish to occu [#permalink] New post   20 Feb 2021, 05:36
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Hello reii1998. out of 4 chairs you can choose 2 chairs in 4C2 ways and then in those 2 chairs, the selected persons can arrange themselves in 2! ways. So the total arrangements = 4C2 * 2! = \(\frac{4!}{2!2!} * 2! = \frac{4!}{2!}\) = 4P2.

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Re: Eight chairs are numbered 1 to 8. Two women and three men wish to occu [#permalink] New post   25 Feb 2021, 11:51
CrackVerbalGMAT wrote:
Hello reii1998. out of 4 chairs you can choose 2 chairs in 4C2 ways and then in those 2 chairs, the selected persons can arrange themselves in 2! ways. So the total arrangements = 4C2 * 2! = \(\frac{4!}{2!2!} * 2! = \frac{4!}{2!}\) = 4P2.

Arun Kumar



You mean that when men selecting the chair that can be done in 1 way because chair are similar and 3 men select 3 chair out of 4 in 1 way ??

Am i correct?
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Re: Eight chairs are numbered 1 to 8. Two women and three men wish to occu [#permalink] New post   28 Feb 2021, 18:26
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jrk23 wrote:
CrackVerbalGMAT wrote:
Hello reii1998. out of 4 chairs you can choose 2 chairs in 4C2 ways and then in those 2 chairs, the selected persons can arrange themselves in 2! ways. So the total arrangements = 4C2 * 2! = \(\frac{4!}{2!2!} * 2! = \frac{4!}{2!}\) = 4P2.

Arun Kumar



You mean that when men selecting the chair that can be done in 1 way because chair are similar and 3 men select 3 chair out of 4 in 1 way ??

Am i correct?



Hello jrk23. After 2 women select 2 seats out of 4, there are still 6 seats remaining. 3 men can select 3 seats in 6C3 ways and then they can rearrange themselves in 3! ways. Therefore for men it is 6C3 * 3! = 6P3


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Re: Eight chairs are numbered 1 to 8. Two women and three men wish to occu [#permalink]
28 Feb 2021, 18:26
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