A 5-digit number is formed using digits 1, 3, 5, 7 and 9 without repeating any one of them. What is the sum of all such possible numbers?
Answers
Step-by-step explanation:
We can form 5! = 120 numbers.
Out of them, each 24 will have 1, 2, 3, 4 & 5 as
thousands, thousands, hundreds, tens & unit di
1 + 2 + 3 + 4 + 5 = 15 so face value of each col
= 15*24 = 360.
Sum = 360 (10,000 + 1,000 + 100 + 10 + 1) =3999960
Answer:
6666600
Step-by-step explanation:
no of all such possible number = 5! = 120
If 1 is fixed in the unit's place, the no of numbers of 5 digits = 4! = 24
Similarly no of numbers with each 3,5,7,9 in the units place = 24
Thus, each digits occurs 24 times in each of units, tens, hundreds, thousands and ten thousands places
The sum of the digits in the unit's place in 120 numbers of 5 digits = 24( 1+3+5+7+9)
= 600
Similarly the sum of digits in the units, tens, hundreds, thousands and ten thousands places = 600
Hence required sum of all the numbers = 600*1 + 600*10 + 600*100 + 600*1000 + 600*10000 = 6666600
HOPE IT HELPS.
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