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Sum of all the numbers that can be formed using all the digits $2,3,3,4,4,4$ is
(A) $22222200$
(B) $11111100$
(C) $55555500$
(D) None of these

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Last updated date: 19th Jun 2024
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Answer
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Hint: For answering this question we will first derive the number of possible ways in which the digits can be arranged to form numbers and then calculate the number of times each digit appears and then sum up all of them.

Complete step-by-step solution
Now considering the question we need to find the sum of all the numbers that can be formed using all the digits $2, 3, 3, 4, 4, 4$.
The number of possible ways in which the digits $2,3,3,4,4,4$ can be arranged is $\dfrac{6!}{2!3!}$. That is simply we can form $60$ numbers using all the digits $2, 3, 3, 4,4, 4$.
Hence, each digit is assumed to be different would occur $\dfrac{60}{6}=10$ times at each of the $6$ places.
As $3$ occurs twice and $4$ occurs thrice, $3$ will occur $2\times 10=20$ times and $4$ will occur $3\times 10=30$ times.
Hence, the sum of the given digits at all places will be $4\times 30+3\times 20+2\times 10=200$.
Hence the required sum of all numbers that can be formed using $2,3,3,4,4,4$ is $\begin{align}
  & 200\times {{10}^{5}}+200\times {{10}^{4}}+200\times {{10}^{3}}+200\times {{10}^{2}}+200\times {{10}^{1}}+200\times {{10}^{0}} \\
 & \Rightarrow 200\left( {{10}^{5}}+{{10}^{4}}+{{10}^{3}}+{{10}^{2}}+{{10}^{1}}+{{10}^{0}} \right) \\
 & \Rightarrow 200\left( 1+10+100+1000+10000+100000 \right) \\
 & \Rightarrow 200\left( 1,11,111 \right) \\
 & \Rightarrow 2,22,22,200 \\
\end{align}$ .
Hence, we can conclude that the sum of all the numbers that can be formed using all the digits $2,3,3,4,4,4$ is $22222200$.
Hence option A is the correct answer.

Note: While answering questions of this type we should be sure with the calculations and simplifications. If we had made a mistake while simplifying this step $200\times {{10}^{5}}+200\times {{10}^{4}}+200\times {{10}^{3}}+200\times {{10}^{2}}+200\times {{10}^{1}}+200\times {{10}^{0}}$ like in case if we had written $200\left( {{10}^{6}}+{{10}^{5}}+{{10}^{4}}+{{10}^{3}}+{{10}^{2}}+{{10}^{1}} \right)$ then we will end up having $\begin{align}
  & \Rightarrow 200\left( 10+100+1000+10000+100000+1000000 \right) \\
 & \Rightarrow 200\left( 11,11,110 \right) \\
 & \Rightarrow 22,22,22,000 \\
\end{align}$.
We can say it is a complete wrong answer.