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NO ONE IS TALLER, BOTH ARE SAME AS A AND B ARE THE SAME PERSON. AS IT IS MENTIONED THAT 500 MEN ARE ARRANGED IN AN ARRAY OF 10 ROWS AND 50 COLUMNS ACCORDING TO THEIR HEIGHTS. LET'S ASSUME THAT POSITION NUMBERS REPRESENT THEIR HEIGHTS. HENCE, THE SHORTEST AMONG THE 50, 100, 150, ... 450, 500 IS PERSON WITH HEIGHT 50 I.E. A. SIMILARLY THE TALLEST AMONG 1, 2, 3, 4, 5, ..... 48, 48, 50 IS PERSON WITH HEIGHT 50 I.E. B. NOW, BOTH A AND B ARE THE PERSON WITH HEIGHT 50. HENCE BOTH ARE THE SAME BILL IS THE QUARTERBACK, KEN IS THE RECEIVER, AND MARK IS THE KICKER. SINCE BILL IS KEN'S FATHER-IN-LAW, BOTH BILL AND KEN MUST BE MARRIED. THIS LEAVES THE ONLY REMAINING PERSON, MARK, TO BE THE BACHELOR (AND HENCE THE KICKER). SINCE BILL IS TALLER THAN THE RECEIVER, BILL MUST NOT BE THE RECEIVER. BY PROCESS OF ELIMINATION, WE CONCLUDE THAT BILL IS THE QUARTERBACK. THE REMAINING POSITION (THE RECEIVER) GOES TO KEN A MUST NOT HAVE SEEN TWO WHITE HATS ON B AND C, OR HE WOULD HAVE KNOWN HIS OWN HAT MUST BE BLACK SINCE THERE ARE ONLY TWO WHITE HATS. SO A'S ANSWER ESTABLISHES THAT AT LEAST ONE OF B OR C'S HAT IS BLACK. BASED ON A'S ANSWER, B KNOWS THAT HE AND C ARE EITHER BOTH WEARING BLACK, OR ONE IS WEARING BLACK AND ONE IS WEARING A WHITE HAT. IF B SEES THAT C IS WEARING A WHITE HAT, THEN HE WOULD KNOW HIS OWN HAT HAD TO BE BLACK. BUT B DOES NOT KNOW WHAT COLOR HAT HE IS WEARING, WHICH MEAN'S C'S HAT IS NOT WHITE AND MUST BE BLACK. SINCE BOTH A AND B CANNOT DEDUCE THE COLOR OF THEIR OWN HATS, C WILL KNOW THAT HE IS WEARING A BLACK HAT FIRST HEAP: 22, SECOND HEAP: 14, THIRD HEAP: 12. TO SOLVE THIS PROBLEM, WE SHALL HAVE TO START FROM THE END. WE HAVE BEEN TOLD THAT AFTER ALL THE TRANSPOSITIONS, THE NUMBER OF MATCHES IN EACH HEAP IS THE SAME. LET US PROCEED FROM THIS FACT. SINCE THE TOTAL NUMBER OF MATCHES HAS NOT CHANGED IN THE PROCESS, AND THE TOTAL NUMBER BEING 48, IT FOLLOWS THAT THERE WERE 16 MATCHES IN EACH HEAP. AND SO, IN THE END WE HAVE: FIRST HEAP: 16, SECOND HEAP: 16, THIRD HEAP: 16 IMMEDIATELY BEFORE THIS WE HAVE ADDED TO THE FIRST HEAP AS MANY MATCHES AS THERE WERE IN IT, I.E. WE HAD DOUBLED THE NUMBER. SO, BEFORE THE FINAL TRANSPOSITION, THERE ARE ONLY 8 MATCHES IN THE FIRST HEAP. NOW, IN THE THIRD HEAP, FROM WHICH WE TOOK THESE 8 MATCHES, THERE WERE: 16 + 8 = 24 MATCHES. WE NOW HAVE THE NUMBERS AS FOLLOWS: FIRST HEAP: 8, SECOND HEAP: 16, THIRD HEAP: 24. WE KNOW THAT WE TOOK FROM THE SECOND HEAP AS MANY MATCHES AS THERE WERE IN THE THIRD HEAP, WHICH MEANS 24 WAS DOUBLE THE ORIGINAL NUMBER. FROM THIS WE KNOW HOW MANY MATCHES WE HAD IN EACH HEAP AFTER THE FIRST TRANSPOSITION: FIRST HEAP: 8, SECOND HEAP: 16 + 12 = 28, THIRD HEAP: 12. NOW WE CAN DRAW THE FINAL CONCLUSION THAT BEFORE THE FIRST TRANSPOSITION THE NUMBER OF MATCHES IN EACH HEAP WAS: FIRST HEAP: 22, SECOND HEAP: 14, THIRD HEAP: 12