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This is a puzzle from Braingle:

Savannah got distracted when her shuttle arrived at Earth Central Airport. Now she's at Lost & Found with four other travelers all waiting to retrieve one lost piece of luggage of a different color (black, blue, green, red or yellow.) While filling out their baggage retrieval paperwork, they discovered an amazing coincidence. Each person had the last name of one other person's nationality and another's destination country (America, England, France, Panama or Zambia). Nobody had the same last name, nationality or destination country of anyone else in the group. Each person is traveling from his or her country of origin to only one other country. Use the following clues to determine each traveler's first name, last name (L), nationality (N), destination country (D) and baggage color. Because this is Logic Land, the unlikelihood of the travelers actually sharing the same airport is not of any concern in solving this puzzle.

  1. Each traveler's last name, nationality and destination country are three different names.
  2. Ms. Panama is neither the Zambian citizen nor the one going to England; Carmelita wants her suitcase to stand out, so she is not the person with the black bag. (All five travelers are mentioned in this clue.)
  3. The person bound for France, who asked Echo where she bought her fragrant cup of coffee, sat next to Mr. England.
  4. Echo is not the French citizen that helped Pierre with his paperwork.
  5. Of the person with the lovely yellow suitcase and Savannah, one is traveling to Zambia and the other is the American citizen.
  6. Pierre's destination country is the nationality of the owner of the black bag; the destination country of the owner of the black bag is Pierre's nationality.
  7. The person surnamed America is not going to France and the person traveling to America has either a blue or green bag.
  8. Juan, who does not have yellow luggage, does not have England as his last name.
  9. The five travelers are the American, Echo, the blue suitcase owner, the one traveling to Panama and Ms. France.

The answer below is wrong, however I am unable to find the error please help:

$\begin{array}{|c|c|c|c|c|}\hline FN&LN&Nationality&Destination&Colour\\\hline SAVANNAH&PANAMA&AMERICAN&FRANCE&BLACK\\ECHO&AMERICA&PANAMA&ZAMBIA&YELLOW\\PIERRE&ENGLAND&FRENCH&AMERICA&BLUE\\JUAN&ZAMBIA&ENGLAND&PANAMA&GREEN/RED\\CARAMELITA&FRANCE&ZAMBIA&ENGLAND&GREEN/RED\\\hline\end{array}$

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  • $\begingroup$ If you would like us to find the error, can you please show us your working out? $\endgroup$ – boboquack Mar 9 '17 at 8:42
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    $\begingroup$ The French citizen helped Pierre with his paperwork, so Pierre can't be the Frenchman. $\endgroup$ – M Oehm Mar 9 '17 at 9:21
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Your solution is wrong in a couple of ways.

Rule #2 says it mentions all five travelers, but your chart shows only three: your Carmelita is the Zambian citizen, and your Ms. Panama has the black bag. Thus in your solution, #2 mentions only Savannah (2x, as Ms. Panama and as black bag owner), Carmelita (2x, as Carmelita and as Zambian), and Juan (going to England).

Also, as pointed out by @M Oehm, Rule #4 precludes Pierre from being French.

The complete solution is this:

FIRST       LAST     NATIONALITY DEST       LUGGAGE
Savannah    France   Panamanian  ZAMBIA     Black
Echo        Panama   English     AMERICA    Green
Pierre      England  Zambian     PANAMA     Red
Juan        America  French      ENGLAND    Blue
Carmelita   Zambia   American    FRANCE     Yellow

You can see the full logic grid at this monster URL:
http://www.jsingler.de/apps/logikloeser/?language=en#(at:s,items:!(!(Savannah,Echo,Pierre,Juan,Carmelita),!(Panama,America,England,Zambia,France),!(Panamanian,American,English,Zambian,French),!(PANAMA,AMERICA,ENGLAND,ZAMBIA,FRANCE),!(Black,Yellow,Blue,Green,Red)),ms:t,n:!(b0d0,b1d1,b0c0,b1c1,b2d2,b3d3,b4d4,b2c2,b3c3,b4c4,b0c3,b0d2,a4b0,a4c3,a4d2,a4e0,b0e0,c3e0,d2e0,c3d2,a1d4,b2d4,a1c4,a2c4,a0e1,c1d3,a2e0,b1d4,d1e0,d1e1,d1e4,a3e1,a3b2,a1c1,a1e2,a1d0,a1b4,c1e2,c1d0,b4c1,d0e2,b4e2,b4d0,a3d3,a3c1,a2d3,a2c1,a2c2,c0d0,c1d1,c2d2,c3d3,c4d4,a1b2,a2b0,a3b0,a0b2,a4b2,a2b4,a3b4,c2e0,c4e0,a2c0,d4e0),nc:5,ni:5,p:!(a0d3,c1e1),v:0)

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