Seven Oscar Nominees are seated at a round table. Three of them have

Question

Seven Oscar Nominees are seated at a round table. Three of them have been known to slap other people for no particular reason. How many different seating arrangements are possible in which none of the three slappers sits next to another slapper?

A. 120
B. 144
C. 720
D. 840
E. 5040

A. 120   
B. 144
C. 720
D. 840 
E. 5040

Gemmie · Answer

Arrangement for the 4 remaining people: (4-1)! = 3! = 6 (because they sit in a circle)

There are 4 spaces (between each 2 of these 4 people) available for the 3 slappers
Pick 3 out of 4 spaces => 4C3 = 4

In each choice (of pick 3 spaces), arrangements of 3 slappers for each choice: 3! = 6

==> Total number of arrangments: 6 * 4 * 6 = 144

NeverTooLateRight · Answer

We have to separate two groups (size 4 and 3) in a table of 7. Since all points are touching (arranged in circle) the only way to do this is to alternate the two groups every time - there can be no consecutive members of the same category.

The circle means it doesn’t matter what category we start with. Let’s say it’s the size 4 group.

We have 4 choices for the first person. 3 choices for the second. 3 for the third. And so on 
4 * 3 * 3 * 2 * 2 * 1 * 1 = 144

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Bunuel · Answer

The number of ways to arrange the remaining 4 nominees around the table is (4 - 1)! = 6. This creates four slots between each pair of non-slappers.

The number of ways to select 3 slots from these 4 is 4C3 = 4.

The 3 slappers can be arranged in these 3 slots in 3! = 6 ways.

Therefore, the total number of ways is 6 * 4 * 6 = 144.

Answer: B.

kop18 · Answer

Hi  Bunuel ,

Why can't I do it this way?

Total arrangements where the 3 do not sit together = Total seatings possible - Total seatings where all 3 sit together
Total seatings possible => (7-1=6)! because in a circular table the no of arrangements possible is (n-1)!
Total seatings where all 3 sit together => (6-3+1=4)!*3! no of arrangements possible if the 3 nominees to be seated together*no of ways the 3 can arrange within themselves

Total = 6! - 4!3! = 720- 24*6 = 720-144= 576.

I know my answer is not in the listed option so I would ideally pick "B" because only that option makes a little sense but please can u suggest where I am going wrong in my working?

Bunuel · Answer

The phrase "none of the three slappers sits next to another slapper" implies that not only can all three slappers not sit together (the cases you accounted for), but also that no two of the three slappers can sit together (the cases you are missing).

kop18 · Answer

Thanks for correcting me  Bunuel 
The moment I read this question I went on to assume that the 3 can't sit together. It did not strick me that the left over 2 also cannot. 
Please can share a list of other such questions in which a similar trap lies please?

Bunuel · Answer

Check this:

Seating Arrangements in a Row and around a Table

Fore more:

21. Combinatorics/Counting Methods 
     Theory   
 Math: Combinatorics 
 Combinatorics Made Easy! 
 1 hour Combinatorics Video Lesson 
 Combinations while dealing with similar and dissimilar things 
 When Permutations & Combinations and Data Sufficiency Come Together 
     Questions   
 Advanced Topic: Probability and Combinations Questions With Solutions 
 Letter Arrangements in Words 
 Seating Arrangements in a Row and around a Table 
 Constructing Numbers, Codes and Passwords 
 DS Questions 
 PS Questions

For even more:
 ALL YOU NEED FOR QUANT ! ! ! 
 Ultimate GMAT Quantitative Megathread

Hope it helps.

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Seven Oscar Nominees are seated at a round table. Three of them have