A five-digit number is formed using digits 1, 3, 5, 8 and 9 without repeating any one of them. What

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14 Aug 2020 Permutation

A five-digit number is formed using digits 1, 3, 5, 8 and 9 without repeating any one of them. What is the sum of all such possible numbers?

OPtion

1) 6666600
2) 6666680
3) 6606600
4) 6066600
5) 6931264
6) 6933264
7) 6932264
8) 6934264
9) 6935264
10) None of these

Solution

If we fix 1 at the unit’s place, the other digits can be arranged in 4! = 24 ways.

So, there are 24 numbers which have 1 at the unit’s place and sum of these numbers is 24.

Similarly, there will be 24 numbers each with 3, 5, 8 and 9 at the unit’s place.

So, the sum of all the numbers at the unit’s place will be:

24 + 24 × 3 + 24 × 5 + 24 × 8 + 24 × 9 = 624

Similarly, the sum of the digits at the ten’s place, hundred’s place,

thousand’s place and ten thousand’s place will be 624.

So, the sum of all numbers = 624 (1 + 10 +100 +1,000 +10,000) = 6933264

Correct Option: 6 Best Solution (5)

M4 Murali   3 years AGO

4!(1+3+5+8+9)=624
624(1+10+100+1000+10000)=6933264

Like? Yes (1) | No   

Bhargav   3 years AGO

Sum of 1,3,5,8,9=26
26(5-1)!=624
so sum =624*11111=6933264

Like? Yes | No (1)  

Devendra Marghade   3 years AGO

Sum of n distinct numbers = (n-1)! * (Sum of all the digits) * (111.. n times)
= 4! * 26 * 11111
= 6933264
Option 6)

Like? Yes | No (1)  

Varun   3 years AGO

Number of permutations = 5P5 = 120
So it is by 120/5 = 24
Unit place will have to be either of 1,3,5,8,9 with 24 repetition
So the sum at unit place = (1+3+5+8+9)*24 = 624
Every place gets same sum
=624
++624
+++624
++++624
+++++624
=6933264

Like? Yes | No (1)  

G.S.Kantharao   3 years AGO

4!*(1+3+5+8+9)*11111
=24*26*11111=6933264.
My option is 6✍️

Like? Yes | No (1)