Question

Find the sum of those four-digit numbers which contain the digits 2,4,5,9 (and the digits are not repeated).

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Answer 1

Sum of all 4 digits number which contains 2,4,5,9(without repetition) = 133320

Explanation:

Let suppose sum of All 4 digits number which contains 2,4,5,9(without repetition) is A×
`10^(3)` +B×
`10^(2)` +C×10 + D(equation 1)

if we fix 2 at unit place ( _ , _ , _ , 2) then total no of possible outcomes is 3!(3×2×1)

and sum of all possible outcomes of which has 2 at unit place is 3!×2

Similarly,

if we fix 4 at unit place then we got sum 3!×4

if we fix 5 at unit place then we got sum 3!×5

if we fix 9 at unit place then we got sum 3!×9

So, the total sum of unit place will be 3!×20

D = 3!×20

In the same above manner we will do for ten's place then we got result

C = 3!×20

In the same above manner we will do for Hundred's place then we got result

B = 3!×20

In the same above manner we will do for Thousand's place then we got result

A = 3!×20

Then, Replace A,B,C & D in equation 1

we got new equation = 3!×20×
`10^(3)` +3!×20×
`10^(2)` +3!×20×10 + 3!×20

lets take 3!×20 common

3!×20(1+
`10^(1)`+
`10^(2)`+
`10^(3)` )

if we observe then it is a Geometric Progression

Sum of "n" number of Geometric Progression is =
`(a_(1)(r^(n)-1 ))/(r-1)`

`a_(1)` = the first term

r = common ratio

n = number of terms

In above case,

`a_(1)` = 1

r = 10

n = 4

hence,
`(1(10^(4)-1 ))/(10-1)` =
`(9999)/(9)` = 1111

Sum of all 4 digits number which contains 2,4,5,9(without repetition) = 3!×20×111

=133320