The sum of all 4 digit number containing the digits 2, 4, 6, 8, without repetitions is
  • a)
    133330
  • b)
    122220
  • c)
    213330
  • d)
    133320
Correct answer is 'D'. Can you explain this answer?

CA Foundation Question

Sandhya Grewal
May 28, 2018
As the 2,4,6,8 creates Arithmetic progression (A.P.) 
THE FORMULA FOR THIS is:

n!/2(first term +last term) 

Using 2,4,6,8: smallest number will be 2468 and this will be first number and 8642 will be the last and largest number.

so, 
the value of n! is 4!=24 
as 4 numbers can be arranged on 4! ways without repetition=24 

Hence,
24/2 (2,468+8,642) 
=12(11,110) 
=133320

Be Positive
Sep 08, 2020
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

Shraddha Patil
Aug 14, 2020
=4!/4(20)(1111)
=24/4(20)(1111)
=6(20)(1111)
=133320

Shubhangi Singhal
Oct 19, 2019
(2+4+6+8) ×3! ×1111 =133320

Nandita Shekhar
Nov 16, 2020
(n-1)! (sum of all digits) (111..... n times)

3! (2+4+6+8) (1111)

6×20*1111

ans :133320

Archa T.a
Dec 01, 2020
Take the sum of all numbers i e, 2+4+6+8=20.
take the average of the sum and it gives =20/4 =5
then, take the average of the sum that is 5 as the 4 digit number then the 4 digit number =5555
and multiply it with 4! that is 4*3*2*1=24.
and then we get 5555*24 which is =133320(option(d))

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