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Three-digit numbers are formed from the digits 1,2 and 3 in such a way that the digits are not repeated. What is the sum of such three-digit numbers?
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The correct answer is:
1332
Sum of the numbers $=$ Sum of given numbers. $(\mathrm{n}-1) !\left[10^{0}+10^{1}+10^{2}+\ldots . .\right]$
Here, sum of th ree digit numbers $=\underline{\text { Sum }}$ of the numbers $(3-1) !\left[10^{\circ}+10^{1}+10^{2}\right]$
$=(1+2+3)(3-1) !\left[10^{0}+10^{1}+10^{2}\right]$
$=6 \times 2 \times 1[1+10+100]$
$=12 \times 111$
$=1332$
Here, sum of th ree digit numbers $=\underline{\text { Sum }}$ of the numbers $(3-1) !\left[10^{\circ}+10^{1}+10^{2}\right]$
$=(1+2+3)(3-1) !\left[10^{0}+10^{1}+10^{2}\right]$
$=6 \times 2 \times 1[1+10+100]$
$=12 \times 111$
$=1332$
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