Question
An eight digit number divisible by 9 is to be formed by using 8 digits out of the digits 0,1,2,3,4,5,6,7,8,9 without replacement.
The number of ways in which can be done is
The number of ways in which can be done is
Answer: Option C
:
C
From the digits 0 1 2 3 4 5 6 7 8 9
The sum is 45
Thus, every time we remove 2 digits which add up to 9, we will have a new set of 8 numbers which are divisible by 9
Case 1
Removing 0 and 9
Number of cases possible = 8!
Case 2
Removing
(1)8 and 1
(2)7 and 2
(3)6 and 3
(4)5 and 4
In each case, total number of cases = 8!-7!
(we subtract 7! Because those are the cases where 0 is the first digit)
Hence, total number of ways = 4(8!-7!)+8!= 36x7!.
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:
C
From the digits 0 1 2 3 4 5 6 7 8 9
The sum is 45
Thus, every time we remove 2 digits which add up to 9, we will have a new set of 8 numbers which are divisible by 9
Case 1
Removing 0 and 9
Number of cases possible = 8!
Case 2
Removing
(1)8 and 1
(2)7 and 2
(3)6 and 3
(4)5 and 4
In each case, total number of cases = 8!-7!
(we subtract 7! Because those are the cases where 0 is the first digit)
Hence, total number of ways = 4(8!-7!)+8!= 36x7!.
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