My Math Forum Deal Or No Deal- How Many Permutations/Combinations?

 August 21st, 2014, 03:59 PM #1 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 673 Thanks: 88 Deal Or No Deal- How Many Permutations/Combinations? https://en.wikipedia.org/wiki/Deal_o...U.S._game_show says there are 26 cases. The First Round removes 6 cases, the Second Round removes 5 cases, the Third Round removes 4 cases, the Fourth Round removes 3 cases, the Fifth Round removes 2 cases, and there are 6 cases removed 1 at a time after that. For the last 6 cases it is just 6! = 720. Each case gets removed exactly once. What round a case is picked in matters but not what order within the round, so the answer will be less than 26! So for the first five rounds how many ways are there? I would multiply that by 720 to get the answer. I'm ignoring the fact that it can be possible to select different combinations of cases that have the same amount of money because dealing with that would make the problem harder. Is the answer: 26 C 6 for the First Round 20 C 5 for the Second round 15 C 4 for the Third Round 11 C 3 for the Fourth Round 8 C 2 for the Fifth Round Multiplying those five numbers by 720 = about 1.62 * 10^19. Is that the answer?

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