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Permutation under Restriction

The numbers 2...

Question

The numbers $2, 3, 4, 5, 6, 7, 8$ are to be placed, one per square, in the diagram shown below such 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 $21.$ Then the number of different ways to do this is

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Solution

As per given condition $a_{1} + a_{2} + a_{3} + a_{4} = 21 \dots (1)$
and
$a_{5} + a_{3} + a_{6} + a_{7} = 21 \dots (2)$
and
$a_{1} + a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7} = 35 \dots (3)$
By equations $(1), (2)$ and $(3)$
$\Rightarrow a_{3} = 7,$
$a_{1} + a_{2} + a_{4} = 14$ and
$a_{5} + a_{6} + a_{7} = 14$

Out of $2, 3, 4, 5, 6, 8$ only $3, 5$ are odd.
So, in order to get sum equals $14,$ number $3$ and $5$ must be in same line.
So, $a_{1}, a_{2}, a_{4} \in {3, 5, 6}$
and $a_{5}, a_{6}, a_{7} \in {2, 4, 8}$
or
$a_{1}, a_{2}, a_{4} \in {2, 4, 8}$ and
$a_{5}, a_{6}, a_{7} \in {3, 5, 6}$

Hence, required number of ways $= (3! \times 3!) + (3! \times 3!) = 72.$

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