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 and 9 without replacement. The number of ways in which this can be done is

Answer 1

firstly, we've to find octets of digits whose sum is a multiple of 9.

As we're given 10 digits, we've to drop 2 digits from them to get octets.

As 0 + .... + 9 = 45 & 45 mod 9 = 0, sum of 2 digits to be dropped must be a multiple of 9 (obviously 9).

It's not a difficult task to find these duplets. Without much trouble, we get 9 = 0+9 = 1+8 = 2+7 = 3+6 = 4+5.

So, there exist such 5 octets. Now, number formed by octet removing (0,8) can reverse itself in 8! ways.

But for other 4 cases, we've to exclude numbers having 0 as 8th digit. So, there are 8!-7! = 7(7!) ways for each case.

We can form 8! + 28(7!)= 36(7!) numbers in this way.

Answer 2

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 and 9 without replacement. The number of ways in which this can be done is