Respuesta :
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×[tex]10^{3}[/tex] +B×[tex]10^{2}[/tex] +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×[tex]10^{3}[/tex] +3!×20×[tex]10^{2}[/tex] +3!×20×10 + 3!×20
lets take 3!×20 common
3!×20(1+[tex]10^{1}[/tex]+[tex]10^{2}[/tex]+[tex]10^{3}[/tex] )
if we observe then it is a Geometric Progression
Sum of "n" number of Geometric Progression is = [tex]\frac{a_{1}(r^{n}-1 )}{r-1}[/tex]
[tex]a_{1}[/tex] = the first term
r = common ratio
n = number of terms
In above case,
[tex]a_{1}[/tex] = 1
r = 10
n = 4
hence,[tex]\frac{1(10^{4}-1 )}{10-1}[/tex] = [tex]\frac{9999}{9}[/tex] = 1111
Sum of all 4 digits number which contains 2,4,5,9(without repetition) = 3!×20×111
=133320
