Question

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. $Noneofthese$

Answer 1

The correct option is A $22222200$

Number of possible arrangements are $\frac{6!}{3!2!}=60$

Hence, each digit if assumed different would occur $\frac{60}{6}=10$ times at each of the 6 places.

Since $4$ occurs $3$ times and $3$ occurs twice, hence at each place $4$ occurs $3\times 10=30$ times and $3$ occurs $10\times 2=20$ times and $2$ occurs $10$ times .

Hence sum of digit at all places will be $30\times 4+20\times 3+10\times 2=200$

Hence sum of all numbers $=200\times 1000000+200\times 100000+200\times 10000+200\times 1000+200\times 100+200\times 10+200$

$=22222200$

Answer-(A)