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Delegates from 10 countries, including Russia, France, England and the United States are to be seated in a row. How many different seating arrangements are possible if the French and English delegates are to be seated next to each other and the Russian and American delegates are not to be next to each other?

The approach I took was first determining the total possible arrangements and then subtracting out the ones that won't work as per the given conditions. So, the total possible arrangments is $10!$. Take the first condition where the French and English delegates must be seated next to each other. Pick a spot for the French delegate. Now, if the French delegate is at one of the end seats, then there is only $1$ spot for the English delegate. This constitutes $2$ possibilities for their seating. Now, if the French delegate is in one of the $8$ interior seats, the English delegate has $2$ possible seats for each one so $16$ pairs. A similar approach would be taken for the Russian and American delegates. Am I heading in the right direction? Any help would be greatly appreciated. Thanks!

RXY15
  • 911

3 Answers3

9

There are total 10 delegates. French and English delegates are to sit together. Considering French and English delegate as a single entity (or another delegate), there are total 9 entities which can be arranged in $9!$ ways. Also French and English delegates can be arranged in $2!=2$ ways: (FE,EF).

So total: $9!\times 2!$

Now its given that US and Russian delegates should not be sited together. To calculate total count with this condition, we calculate total ways in which US and Russian delegates sit together and subtract it from above.

With French and English delegates sited together, we have total 9 entities. Out of these, now we want to find in how many ways US and Russian delegates can sit together. Following same above approach. we interpret US and Russian delegate as single entity. So we are now having 8 entities which can be arranged in $8!$ ways. US and Russian delegates can be arranged in $2!$ ways among themselves: (UR,RU). Also considering French and English delegates can be arranged in $2!=2$ ways: (FE,EF).

So total: $8!\times 2!\times 2!$

Final answer is obtained by subtracting number of arrangements in which both

  • US-Russian sit together and
  • French-English sit together

from number of arrangements in which French-English sit together:

$9!\times 2! - 8! \times 2! \times 2! = 564480$

Notice this follows principle of inclusion-exclusion.

Another approach

French and English delegates are to sit together. Considering French and English delegate as a single entity (or another delegate), there are total 9 entities. Also French and English delegates can be arranged in $2!=2$ ways: (FE,EF).

Now out of those 9 entities, American and Russian delegates should not seat next to each other. So we first permute remaining 7 entities (6 delegates plus one entity containing French and English delegate) in $7!$ ways.

These 7 entities will have 8 gaps between them which can be occupied by American and Russian delegates, so that they wont seat next to each other. This can be done in ${}^8P_2$ ways.

So the total count: $2!\times 7!\times {}^8P_2=564480$

Mahesha999
  • 2,053
2

First solve the problem for $9$ instead of $10$ delegates, in the sense that the French and English "go together". The outcome of it must be multiplied by factor $2$ because there are two orders possible: $FE$ and $EF$.

Place Russia. In the adapted problem there are $2+7=9$ possibilities. The $2$ stands for places completely on the right and completely on the left and the $7$ stands for the other places.

Now place the US. If Russia is placed completely on the left or completely on the right then there are $7$ possibilities for the US and in the other cases there are $6$.

Now $7$ spots are left for the rest leading to $7!$ possibilities.

So finally we come to $2\times[2\times7+7\times6]\times7!$ possibilities.

drhab
  • 151,093
0

This is my approach:

2!* 9 * 8! - 2! *2! * 8 * 7!

(2! => number of ways French and English delegates arrange them-self. so they are single entity )

(9 => number of ways French and English delegates can sit from remaining 8 places )

ex: 

    F E _ _ _ _ _ _ _ _

    _ F E _ _ _ _ _ _ _

    _ _ _ F E _ _ _ _ _ 

 and so on  

    _ _ _ _ _ _ _ _ F E

(8! => other 8 delegates arrangement possibilities )

Now another condition is US and Russian delegates not sit together. We can calculate by subtracting number of ways they sit together from the above ans.

(2! => number of ways US and Russian arrange them-self. so they are single entity)

(2! => number of ways French and English delegates arrange them-self. so they are single entity)

(8 => number of ways US and Russian delegates can sit from remaining 7 places )

ex: 

U R _ _ _ _ _ _ F E

_ U R _ _ _ _ _ F E

and so on 


F E _ _ _ _ _ _ U R

(7! => number of ways remaining 7 arrange them-self. Note: French and English are single entity )

loyola
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