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2022/23 AHSMC Part 1 Problem 13

Problem 13 of 16

In how many ways can we place six letters AA, AA, BB, BB, CC, CC in the grid below, such that each square contains exactly one letter and no column contains 22 identical letters?

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Solution

There are (62)−3=12\dbinom{6}{2} - 3 = 12 acceptable choices to place the letters AA, AA in the grid while for the other letters BB, BB, CC, CC remain (42)−2=4\dbinom{4}{2} - 2 = 4 acceptable choices. Hence the total number of acceptable choices is 12⋅4=4812 \cdot 4 = 48. The answer is (C).

Alternative solution: There are (62)⋅(42)=15⋅6=90\dbinom{6}{2} \cdot \dbinom{4}{2} = 15 \cdot 6 = 90 ways of putting the letters into the grid with no restriction. We get an unacceptable result in one of the following two situations:

  1. choose a column (33 ways) and put the same letter in both places in this column (33 ways), then complete the rest of the grid using the other two letters so that both remaining columns get both letters ((42)−2=4\dbinom{4}{2} - 2 = 4 ways); thus there are 3⋅3⋅4=363 \cdot 3 \cdot 4 = 36 ways of filling the grid so that there is exactly one column containing identical letters.
  2. all three columns have identical letters (3!=63! = 6 ways).

Therefore the total number of acceptable solutions is 90−36−6=4890 - 36 - 6 = 48.

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