2019 Mock AMC 10B Problems/Problem 21

Solution 1: Author:(Shiva Kannan)

There are $4! = 24$ ways to place one of each $1, 2, 3,$ and $4$ in the top row. For the second row, we must count all the possible arrangements other than the arrangements in which the value in the first row and a particular column and the value in the second row and that same column, never are the same. The total number of ways regardless of the condition is $24$ . We can use PIE. $Number of arrangements in which there is a$ 1 $under the$ 1 $in the first row is$ 3! = 6 $, as the remaining three numbers of the row can be arranged in$ 6 $ways. By symmetry, this is the same for all$ 4 $numbers, so the total number of cases in which one value is below itself is$ 24 $. But we have to correct for overcounting. Number of arrangements in which both$ 1 $and$ 2 $are below themselves is$ 2! = 2 $, as there are$ 2 $ways to place the remaining two elements. Like this, we can take any pair of numbers from the four given numbers, so the total number of cases in which$ 2 $values are below themselves is$ 2* {4 \choose 2} = 12 $. For$ 3 $numbers being under themselves, we can place the remaining number in$ 1 $way, and there are three possible sets of$ 3 $numbers from the four to pick, so the number of cases in which$ 3 $numbers are below themselves is$ 4 $. For all four numbers being under themselves, there is only$ 1 $case.$ 24 - 24 + 12 - 4 + 1 = 9 $, which yields$ 9$ ways to construct the second row.

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2019 Mock AMC 10B Problems/Problem 21

Solution 1: Author:(Shiva Kannan)