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# In the figure, two four digit numbers are to be formed by filling the places with digits. The number of different ways in which the places can be filled by the digits so that the sum of the numbers formed is also a four-digit number and in no place there is addition with carrying, is equal to (a). ${{55}^{6}}$(b). 220(c). ${{45}^{4}}$(d). 219

Last updated date: 24th Jun 2024
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Hint: In this question, we are given to find the digits so that there will be no carrying over of the digits at any place and that the sum should also be a 4-digit number. Therefore, we can see that for the condition of no carrying over, the sum of the digits can be at most nine in a given column. Also, as the sum of the numbers has to be a four-digit number, the thousands place cannot be equal to zero.

\begin{align} & \sum\limits_{x=0}^{x=9}{10-x}=\left( 10-0 \right)+\left( 10-1 \right)+\left( 10-2 \right)+...+\left( 10-9 \right) \\ & =10\times 10-\left( 0+1+....+9 \right)=100-\dfrac{9\times 10}{2}=55...............(1.1) \\ \end{align}
Where we have used the fact that the sum of first n natural numbers is equal to $\dfrac{n(n+1)}{2}$.