The number of ways to fill each of the four cells of the table with a distinct natural number, such that the sum of the numbers is $ 10 $ and the sum of the numbers places diagonally are equal is

Question

The number of ways to fill each of the four cells of the table with a distinct natural number, such that the sum of the numbers is $ 10 $ and the sum of the numbers places diagonally are equal is, $ 2 ! \times 2 ! \times 2 ! $, $ 4 ! $, $ 2(4 !) $, $ 2 ! $

MARKS App · Accepted Answer

The natural numbers are 1 , 2 , 3 and 4 (any others would lead to the sum exceeding 10 ) Clearly, in one diagonal we have to place 1 , 4 And in the other 2 , 3 So we first select a diagonal, and then we arrange the numbers in its boxes Number of ways of selection of the diagonal = 2 ! The number of arrangements of numbers = 2 ! × 2 ! × 2 ! = 8 .

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