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

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In how many ways can Ann, Bob, Chuck, Don and Ed be seated [#permalink] New post 08 Oct 2017, 22:28
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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] New post 08 Oct 2017, 23:01
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Bunuel 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

Kudos for correct solution.



There are 5 people

so it can be arranged in 5! ways i.e = 120 ways

Now ques has asked Ann and Bob cannot sit next to each other.

So we need to find out the ways it can sit together and then substract by total no. of ways i.e 5!

So Ann and bob can sit next to each other = 4! (This is because we have taken Ann and Bob as one and number of ways these 4 people can arrange is 4!)

But Ann and Bob can also can arrange themselves in 2! ways.

Therefore total no. ways Ann and Bob can arrange themselves = 4!*2!

Thus,

The number of ways Ann and bob cannot be seated next to each other = 5! - (4! *2!) = 72
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Re: In how many ways can Ann, Bob, Chuck, Don and Ed be seated [#permalink] New post 13 Dec 2017, 15:45
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Bunuel 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.

Answer: D
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Re: In how many ways can Ann, Bob, Chuck, Don and Ed be seated [#permalink] New post 13 Dec 2017, 18:45
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Let's focus on the restrition and as follows
# number of arrangements in which Ann and Bob are not seated next to each other = # total of arrangements - # total of arrangements in which they are seated to next each other
We can simply count the least one : 8 arrangements.
And they are 4 seats left which lead to 3 * 2 * 1 = 6 possibilities to place the four other if Ann and Bob were required to seat to next each other.
We simply multiply both 8 * 6 = 48.

The total number of arrangements can be counted as follows = 5! Which is 120

Then we plug the values in our equation.
# number of arrangements when they are NOT next to each other = total arrangement - total arrangement if they were next to each other
120 -48 = 72
D
Re: In how many ways can Ann, Bob, Chuck, Don and Ed be seated   [#permalink] 13 Dec 2017, 18:45
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In how many ways can Ann, Bob, Chuck, Don and Ed be seated

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