JEE MAIN - Mathematics (2021 - 18th March Morning Shift - No. 14)
The sum of all the 4-digit distinct numbers that can be formed with the digits 1, 2, 2 and 3 is :
26664
122664
122234
22264
Explanation
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
= $${{4!} \over {2!}}$$ = 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
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