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There are 9 letters having different colors (red, orange, yellow, green, blue, indigo, violet) and 4 boxes each of different shapes (tetrahedron, cube, polyhedron, dodecahedron). How many ways are there to place these 9 letters into the 4 boxes such that each box contains at least 1 letter?(a) 260100(b) 878760(c) 437102(d) 256850The question was posed to me in examination.My enquiry is from Discrete Probability topic in division Discrete Probability of Discrete Mathematics

Answer»

The correct answer is (a) 260100

To elaborate: Let N be the total number of ways we can distribute the letters. Each letter can be placed into any one of the 4 BOXES, so |N| = 4^9. Let T be the SET of ways such that the tetrahedron box has no letters, C be the set of ways such that the cube box has no letters, P be the set of ways such that the cube box has no letters, and D be the set of ways such that the dodecahedron box has no letters. Now, to find |N| – |T U C U P U D|. We have |T|=|C|=|P|=|D|=2^7 and SINCE the letters can be placed into one of the two other boxes, and |TUC| = |C U P| = |P U D| = |D U T| = 1^7, since all the letters must be placed in the remaining box, andT ⋂ C ⋂ P ⋂ D| = 0. Hence, PIE implies |N| – |T U C U P U D| = 4^9 – 4 x 2^9 + 4 x 1^9 – 0 = 260100.


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