The numbers 2, 3, 4, 5, 6, 7, 8 are to be placed, one per square, in the diagram shown so that the sum of the four numbers in the horizontal row equals 21 and the sum of the four numbers in the vertical column also equals 21. In how many different ways can this be done?

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answer is 72.

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Detailed Solution

$(a_{1} + a_{2} + a_{3} + a_{4}) = 21$

$a_{5} + a_{3} + a_{6} + a_{7} = 21$

$a_{1} + a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7} = 35$

$\Rightarrow a_{3} = 7, a_{1} + a_{2} + a_{4} = 14, a_{5} + a_{6} + a_{7} = 14$

$a_{1}, a_{2}, a_{4} \in {2, 4, 8}, a_{5}, a_{6}, a_{7} \in {3, 5, 6}$

$a_{1}, a_{2}, a_{4} \in {3, 5, 6}, a_{5}, a_{6}, a_{7} \in {2, 4, 8}$

Required number of ways $= (3! \times 3!) + (3! \times 3!) = 72$

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