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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 a1+a2+a3+a4=21 (1)
and
a5+a3+a6+a7=21 (2)
and
a1+a2+a3+a4+a5+a6+a7=35 (3)
By equations (1),(2) and (3)
a3=7,
a1+a2+a4=14 and
a5+a6+a7=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, a1,a2,a4{3,5,6}
and a5,a6,a7{2,4,8}
or
a1,a2,a4{2,4,8} and
​​​​​​​a5,a6,a7{3,5,6}

Hence, required number of ways =(3!×3!)+(3!×3!)=72.

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