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The sum of all 4-digit numbers that can be formed by using the digits $2,4,6,8$ (repetition of digits is not allowed) is
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Verified Answer
The correct answer is:
133320
There are $4 !=24$ numbers. Each digit occurring $3 !=6$ times, in the unit's, ten's, hundred's and thousand's places. We note that $(2+4+6+8)=120$. Thus in the over all sum there will be 120 units, 120 tens, 120 hundreds and 120 thousands.
The required sum
$\begin{aligned}
&=120\left(1+10+10^{2}+10^{3}\right) \\
&=120 \times 1111=133320
\end{aligned}$
The required sum
$\begin{aligned}
&=120\left(1+10+10^{2}+10^{3}\right) \\
&=120 \times 1111=133320
\end{aligned}$
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