Upar diye gye prashn ka sabse aasaan trika
pehle to samjhte hain ki prashn kehna kya chahta hain
usme (2,4,6,8) 4 alag alag anukramank diye gye hain uske jitne sankhya ban skte hain banana hai phir sabko jod dena hain...,
1111 x (2+4+6+8) x 4!/4 ----> ye aapka aasan trika hain
1111 ye kramank honge utne hi digit lene hain
(2+4+6+8) ye sabka sum hai yaani diye gye 4 kramank ko jod dijiye
4!/4 kitne no. ki possibility hain usse n se divide kr dijiye

1111 x 20 x 6
1,33,320

Given, the digits are 2, 4, 6, 8 without repetitions.

To find the sum of all 4 digit numbers containing these digits, we need to find the sum of each digit in each place (thousands, hundreds, tens, and units) and then sum them up.

Let's consider each place value separately:

Thousand's place:
We have four options for the thousands place: 2, 4, 6, or 8.
Each of these digits will appear in the thousands place 6 times (since there are 3 options for the hundreds place, 2 options for the tens place, and 1 option for the units place).
So the sum of the digits in the thousands place will be:

(2 + 4 + 6 + 8) x 6 x 1000 = 12000

Hundred's place:
We have three options for the hundreds place (since we can't use the digit we used in the thousands place).
Each of these digits will appear in the hundreds place 6 times (since there are 2 options for the tens place and 1 option for the units place).
So the sum of the digits in the hundreds place will be:

(2 + 4 + 6 + 8) x 6 x 100 = 1200

Ten's place:
We have two options for the tens place (since we can't use the digits we used in the thousands and hundreds places).
Each of these digits will appear in the tens place 6 times (since there is 1 option for the units place).
So the sum of the digits in the tens place will be:

(2 + 4 + 6 + 8) x 6 x 10 = 240

Unit's place:
We have one option for the units place (since we can't use the digits we used in the thousands, hundreds, and tens places).
Each of these digits will appear in the units place 6 times.
So the sum of the digits in the units place will be:

(2 + 4 + 6 + 8) x 6 = 120

Adding up the sums from each place value:

12000 + 1200 + 240 + 120 = 13320

Therefore, the correct option is (D) 133320.