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YOU WOULD ONLY NEED TO TAKE OUT ONE MARBLE BECAUSE WE KNOW THAT ALL OF THE LABELS ARE INCORRECT. SO YOU PULL ONE MARBLE OUT OF THE BOX LABELED "MIXED." IF RED COMES OUT, YOU KNOW THAT HAS TO BE THE ALL-RED BOX, SO YOU PUT THE RED LABEL ON IT. THE BOX LABELLED "BLUE" MUST THEN BE LABELLED "MIXED" BECAUSE YOU KNOW IT IS ALSO LABELED INCORRECTLY, AND THEREFORE CAN'T BE BLUE. YOU WOULD LABEL THE LAST BOX "BLUE" BECAUSE THAT IS THE ONLY COLOR/BOX COMBO LEFT. IF THE FIRST MARBLE YOU PULLED OUT FROM THE BOX LABELED "MIXED" IS A BLUE MARBLE, THEN YOU SOLVE THE PROBLEM IN THE SAME GENERAL WAY THE SWITCH WILL BE IN THE OFF POSITION. AFTER SWITCHING 4 TIMES, THE SWITCH WILL RETURN TO ITS ORIGINAL POSITION. SO, 3922 / 4 = 980 WITH REMAINDER 2. THE SWITCH WILL BE AT MEDIUM POSITION AFTER SWITCHING 3920 TIMES, AND AT OFF POSITION AFTER ANOTHER 2 SWITCHES 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