Question 161 of 278MediumMCQ
The sum of all the 4-digit distinct numbers that can be formed with the digits 1, 2, 2 and 3 is:
Solutions · 1
Official solution
Total possible numbers using 1, 2, 2 and 3 is
= \[{{4!} \over {2!}}\] = 12
When unit place is 1, the total possible numbers using remaining 2, 2 and 3 are
= \[{{3!} \over {2!}}\] = 3
When unit place is 2, the total possible numbers using remaining 1, 2 and 3 are
= 3! = 6
When unit place is 3, the total possible numbers using remaining 1, 2 and 2 are
= \[{{3!} \over {2!}}\] = 3
\[ \therefore \] Sum of unit places of all (3 + 6 + 3) 12 numbers is
= ( 1\[ \times \]3 + 2\[ \times \]6 + 3\[ \times \]3)
Similarly,
When 10th place is 1, the total possible numbers using remaining 2, 2 and 3 are
= \[{{3!} \over {2!}}\] = 3
When 10th place is 2, the total possible numbers using remaining 1, 2 and 3 are
= 3! = 6
When 10th place is 3, the total possible numbers using remaining 1, 2 and 2 are
= \[{{3!} \over {2!}}\] = 3
\[ \therefore \] Sum of 10th places of all (3 + 6 + 3) 12 numbers is
= ( 1\[ \times \]3 + 2\[ \times \]6 + 3\[ \times \]3) \[ \times \] 10
Similarly,
Sum of 100th places of all (3 + 6 + 3) 12 numbers is
= ( 1\[ \times \]3 + 2\[ \times \]6 + 3\[ \times \]3) \[ \times \] 100
and Sum of 1000th places of all (3 + 6 + 3) 12 numbers is
= ( 1\[ \times \]3 + 2\[ \times \]6 + 3\[ \times \]3) \[ \times \] 1000
\[ \therefore \] Total sum = ( 1\[ \times \]3 + 2\[ \times \]6 + 3\[ \times \]3) + ( 1\[ \times \]3 + 2\[ \times \]6 + 3\[ \times \]3) \[ \times \] 10
+ ( 1\[ \times \]3 + 2\[ \times \]6 + 3\[ \times \]3) \[ \times \] 100 + ( 1\[ \times \]3 + 2\[ \times \]6 + 3\[ \times \]3) \[ \times \] 1000
= (3 + 12 + 9) (1 + 10 + 100 + 1000) = 1111 \[ \times \] 24 = 26664
Answer the question to unlock