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# In how many ways can Ann, Bob, Chuck, Don and Ed be seated

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Senior Manager
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Joined: 15 Sep 2010
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In how many ways can Ann, Bob, Chuck, Don and Ed be seated  [#permalink]

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25 Nov 2010, 19:06
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45% (medium)

Question Stats:

61% (01:23) correct 39% (01:29) wrong based on 296 sessions

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In how many ways can Ann, Bob, Chuck, Don and Ed be seated in a row such that Ann and Bob are not seated next to each other?

A. 24
B. 48
C. 56
D. 72
E. 96

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Re: In how many ways can Ann, Bob, Chuck, Don and Ed be seated  [#permalink]

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25 Nov 2010, 21:55
4
6
shrive555 wrote:
In how many ways can Ann, Bob, Chuck, Don and Ed be seated in a row such that Ann and Bob are not seated next to each other?

24
48
56
72
96

When the constraint on an arrangement says, "Two people should not be seated together," we do just the opposite. We make them sit together! Kind of tie them with a thread and assume they are one unit!
Let's see why....

These 5 people can be arranged in 5! ways. These are the total number of ways you get.
Now, when we tie 2 people together, we have only 4 entities to arrange. We can do this in 4! ways. But in each of these entities, the two people can sit in two different ways (AB and BA). So number of ways in which these two people sit together is 4!*2!.

Now, the ways in which these two people will not be together will be 5!- 4!*2! = 4!(5 - 2) = 72
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Karishma
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Re: In how many ways can Ann, Bob, Chuck, Don and Ed be seated  [#permalink]

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26 Nov 2010, 13:22
my solution:

3! * 6*2 = 72

3! for Chuck, Don and Ed

6 positions for Ann and Bob: AC AD AE BC BE CE * 2 (as they can switch they positions)

It is not as concise as Karishma's but it works.
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Re: In how many ways can Ann, Bob, Chuck, Don and Ed be seated  [#permalink]

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27 Nov 2010, 14:13
craky wrote:
my solution:

3! * 6*2 = 72

3! for Chuck, Don and Ed

6 positions for Ann and Bob: AC AD AE BC BE CE * 2 (as they can switch they positions)

It is not as concise as Karishma's but it works.

Yes, your logic is absolutely fine. If the number of people is manageable, we can directly find the number of ways in which the two of them should not be seated together.
You arranged C, D and E in 3! ways.
Attachment:

Ques2.jpg [ 3.56 KiB | Viewed 7573 times ]

The 4 dots show 4 positions for Ann.When Ann occupies one of these positions, 3 positions are left over for Bob. So Ann and Bob can sit in 4*3 = 12 ways.
Total 3! * 12 = 72 ways.
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Re: In how many ways can Ann, Bob, Chuck, Don and Ed be seated  [#permalink]

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13 Oct 2015, 20:24
tuanquang269 wrote:
In how many ways can Ann, Bob, Chuck, Don and Ed be seated in a row such that Ann and Bob are not seated next to each other?

(A) 24

(B) 58

(C) 56

(D) 72

(E) 96

In these types of questions, always try to subtract the not allowed arrangements from the total arrangements. This makes our lives much easier.
Here we will assume that Ann and Bob always sit together i.e. they are a single entity
Total arrangements: 5!

Arrangements in which Ann and Bob sit together = 4!*2!
We get this because, total arrangements =4! and Ann and Bob can be arranged in 2! ways

Subtracting this from the total: 5! - 4!*2! = 72. Option D
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Re: In how many ways can Ann, Bob, Chuck, Don and Ed be seated  [#permalink]

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19 Apr 2017, 16:02
1
1
shrive555 wrote:
In how many ways can Ann, Bob, Chuck, Don and Ed be seated in a row such that Ann and Bob are not seated next to each other?

A. 24
B. 48
C. 56
D. 72
E. 96

We can use the following formula:

Total number of ways to arrange the 5 people = (number of arrangements when Bob sits next to Ann) + (number of arrangements when Bob does not sit next to Ann)

Let’s determine the number of arrangements when Bob sits next to Ann.

We can denote Ann, Bob, Chuck, Don, and Ed as A, B, C, D, and E, respectively.

If Ann and Bob must sit together, we can consider them as one person [AB]. For example, one seating arrangement could be [AB][C][D][E]. Thus, the number of ways to arrange four people in a row is 4! = 24.

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

Therefore, the total number of seating arrangements is 24 x 2 = 48 if Ann and Bob DO sit next to each other.

Since there are 5 people being arranged, the total number of possible arrangements is 5! = 120.

Thus, the number of arrangements when Bob does NOT sit next to Ann is 120 - 48 = 72.

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Re: In how many ways can Ann, Bob, Chuck, Don and Ed be seated  [#permalink]

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23 Nov 2018, 21:08
This is a classic FCP with restrictions problem. It's faster to use the "complement" to solve these problems than to count the arrangements directly. If we can count the number of total arrangements, then subtract the number of cases we don't want, we're left with the number of cases we do want.

[# of arrangements with Ann & Bob not sitting together]

= [# without restrictions] - [# with Ann & Bob sitting together]

When arranging 5 people where order matters, the # of arrangements is 5! = 120.

To calculate the # of arrangements where Ann & Bob are seated together, we "glue" them together as a single unit: [Ann, Bob]. In that case we're arranging 4 units where order matters, that's 4! = 24.

But wait! Ann, Bob, Chuck, Don, Ed is different from Bob, Ann, Chuck, Don, Ed even though Ann and Bob are still seated together before the rest. So, there's an additional 24 arrangements where we have them glued together as [Bob, Ann].

The # of arrangements with Ann & Bob not sitting together is 120 - 24 - 24 = 72.

For a review, on this "complement" concept check out the FCP with Restrictions lesson video.
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Re: In how many ways can Ann, Bob, Chuck, Don and Ed be seated   [#permalink] 23 Nov 2018, 21:08
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