Question

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

Answer 1

Answer:

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

Step-by-step 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 = $\frac{a_{1}(r^{n}-1 )}{r-1}$

$a_{1}$ = the first term

r = common ratio

n = number of terms

In above case,

$a_{1}$  = 1

r = 10

n = 4

hence,$\frac{1(10^{4}-1 )}{10-1}$ = $\frac{9999}{9}$ = 1111

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

=133320