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In how many ways can Ann, Bea, Cam, Don, Ella and Fey be seated if Ann

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In how many ways can Ann, Bea, Cam, Don, Ella and Fey be seated if Ann  [#permalink]

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New post 21 Jan 2018, 03:25
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A
B
C
D
E

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Re: In how many ways can Ann, Bea, Cam, Don, Ella and Fey be seated if Ann  [#permalink]

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New post 21 Jan 2018, 04:59
C should be.
Total - AB sit together
total =6!
AB sit together = 2*5!
hence 6!-2*5!= 5!(6-2)= 120*4 = 480
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Re: In how many ways can Ann, Bea, Cam, Don, Ella and Fey be seated if Ann  [#permalink]

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New post 21 Jan 2018, 05:22
What if it's a circular table?
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Re: In how many ways can Ann, Bea, Cam, Don, Ella and Fey be seated if Ann  [#permalink]

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New post 21 Jan 2018, 07:16
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Re: In how many ways can Ann, Bea, Cam, Don, Ella and Fey be seated if Ann  [#permalink]

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New post 23 Jan 2018, 00:15
Bunuel wrote:
In how many ways can Ann, Bea, Cam, Don, Ella and Fey be seated if Ann and Bea cannot be seated next to each other?

(A) 240
(B) 360
(C) 480
(D) 600
(E) 720


The total number of ways all can be seated is 6! & the number of ways Ann & Bea can be seated together is 2*5!.
So total would be 6! - 2*5! = 480.
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Re: In how many ways can Ann, Bea, Cam, Don, Ella and Fey be seated if Ann  [#permalink]

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New post 29 Jan 2018, 06:58
Hi,

I understand the 6! but could someone please explain why it's 2*5! and not 6C2 for the number of ways Ann & Bea can seat next to each other ? :oops:
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Re: In how many ways can Ann, Bea, Cam, Don, Ella and Fey be seated if Ann  [#permalink]

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New post 29 Jan 2018, 07:02
tahina wrote:
Hi,

I understand the 6! but could someone please explain why it's 2*5! and not 6C2 for the number of ways Ann & Bea can seat next to each other ? :oops:


Consider AB as 1 unit, now total 5 ppl are to be seated hence 5! ways; but AB themselves can be seated in 2! i.e. 2 ways, hence 2*5!
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Re: In how many ways can Ann, Bea, Cam, Don, Ella and Fey be seated if Ann  [#permalink]

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New post 30 Jan 2018, 00:31
why do you subtract

"AB sit together = 2*5!"

2*5! here?
how do you come up with These numbers
thank you
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Re: In how many ways can Ann, Bea, Cam, Don, Ella and Fey be seated if Ann  [#permalink]

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New post 30 Jan 2018, 00:40
Jannnn04 wrote:
why do you subtract

"AB sit together = 2*5!"

2*5! here?
how do you come up with These numbers
thank you


In how many ways can Ann, Bea, Cam, Don, Ella and Fey be seated if Ann and Bea cannot be seated next to each other?

(A) 240
(B) 360
(C) 480
(D) 600
(E) 720

6 people (Ann, Bea, Cam, Don, Ella and Fey) without any restriction can be seated in a row in 6! ways.

Now, consider as one unit {Ann, Bea}. We'll have 5 units: {Ann, Bea}; {Cam}; {Don}; {Ella} and {Fey}. These 5 units can be arranged in 5! ways. Ann and Bea within their unit can be arranged in two ways: {Ann, Bea} or {Bea, Ann}. So, the number of ways to arrange 6 people so that two of them, {Ann, Bea}, ARE together is 5!*2.

Finally, if we subtract the number of arrangements where Ann and Bea are together from total number of arrangements, we'll get the number of arrangements where Ann and Bea are NOT together: 6! - 5!*2 = 480.

Answer: C.
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Re: In how many ways can Ann, Bea, Cam, Don, Ella and Fey be seated if Ann  [#permalink]

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New post 30 Jan 2018, 00:42
Bunuel wrote:
In how many ways can Ann, Bea, Cam, Don, Ella and Fey be seated if Ann and Bea cannot be seated next to each other?

(A) 240
(B) 360
(C) 480
(D) 600
(E) 720


Check other Arrangements in a Row and around a Table questions in our Special Questions Directory.
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Re: In how many ways can Ann, Bea, Cam, Don, Ella and Fey be seated if Ann  [#permalink]

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New post 31 Jan 2018, 00:28
Bunuel wrote:
Jannnn04 wrote:
why do you subtract

"AB sit together = 2*5!"

2*5! here?
how do you come up with These numbers
thank you


In how many ways can Ann, Bea, Cam, Don, Ella and Fey be seated if Ann and Bea cannot be seated next to each other?

(A) 240
(B) 360
(C) 480
(D) 600
(E) 720

6 people (Ann, Bea, Cam, Don, Ella and Fey) without any restriction can be seated in a row in 6! ways.

Now, consider as one unit {Ann, Bea}. We'll have 5 units: {Ann, Bea}; {Cam}; {Don}; {Ella} and {Fey}. These 5 units can be arranged in 5! ways. Ann and Bea within their unit can be arranged in two ways: {Ann, Bea} or {Bea, Ann}. So, the number of ways to arrange 6 people so that two of them, {Ann, Bea}, ARE together is 5!*2.

Finally, if we subtract the number of arrangements where Ann and Bea are together from total number of arrangements, we'll get the number of arrangements where Ann and Bea are NOT together: 6! - 5!*2 = 480.

Answer: C.



Ok thank you, much clearer !
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Re: In how many ways can Ann, Bea, Cam, Don, Ella and Fey be seated if Ann  [#permalink]

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New post 31 Jan 2018, 03:48
Bunuel wrote:
In how many ways can Ann, Bea, Cam, Don, Ella and Fey be seated if Ann and Bea cannot be seated next to each other?

(A) 240
(B) 360
(C) 480
(D) 600
(E) 720


Solution



    • We are given the names of 6 people. Let me denote them as A, B, C, D, E and F.
    • We need to find the number of ways in which these 6 people can be arranged if A and B do not sit together.
    • Now, there could be a number of cases in which A and B can be placed away from each other. There could one person between them or two or three or more than that. Finding each and every case would be difficult.
    • Thus, the best approach is to find all the cases of arrangement and subtract from it, the cases in which A and B are always together.
      o The number of ways in which these 6 people can be arranged = 6P6 = 6! = 720
      o The number of ways in which these 6 can be arranged with A and B always together = 5! * 2! = 240
         Since A and B are together we will consider them as one entity and there are remaining 4 people, so there are total 5 entities and since A and B can arrange among themselves that is why, I have multiplied it by 2!
    • Therefore, the number of way in which these 6 people can be arranged and A and B are NOT seated together = 720 – 240 = 480.

Thus, the correct answer is Option C

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Re: In how many ways can Ann, Bea, Cam, Don, Ella and Fey be seated if Ann  [#permalink]

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New post 31 Jan 2018, 17:11
Bunuel wrote:
In how many ways can Ann, Bea, Cam, Don, Ella and Fey be seated if Ann and Bea cannot be seated next to each other?

(A) 240
(B) 360
(C) 480
(D) 600
(E) 720


We can use the following formula:

Total number of ways to arrange the 6 people = (number of arrangements in which Bea sits next to Ann) + (number of arrangements in which Bea does not sit next to Ann)

Let’s determine the number of arrangements in which Bea sits next to Ann.

We can denote Ann, Bea, Cam, Don, Ella, and Fey as A, B, C, D, E, and F respectively.

If Ann and Bea must sit together, we can consider them as one person [AB]. For example, one seating arrangement could be [AB][C][D][E][F]. Thus, the number of ways to arrange five people in a row is 5! = 120.

However, we must also account for the ways we can arrange Ann and Bea, that is, either [AB] or [BA]. Thus, there are 2! = 2 ways to arrange Ann and Bea.

Therefore, the total number of seating arrangements is 120 x 2 = 240 if Ann and Bea DO sit next to each other.

Since there are 6 people being arranged, the total number of possible arrangements is 6! = 720.

Thus, the number of arrangements in which Bea does NOT sit next to Ann is 720 - 240 = 480.

Answer: C
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Re: In how many ways can Ann, Bea, Cam, Don, Ella and Fey be seated if Ann &nbs [#permalink] 31 Jan 2018, 17:11
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