Five digit numbers divisible by 9 are to be formed by using the digits 0, 1, 2, 3, 4, 7, 8 (without repetition The total number of such numbers is equal to

We know that if a number is divisible by 9, then sum, the digits must be divisible by 9. If we follow then the method as in last question we will have ${ }^7{C}_5=\frac{7.6}{12}=21$ Out of these 21 selections we will choose those whose sum is divisible by 9. Il will be very lengthy, so ' choose the alternative method. Sum of given digits =0+1+2+3+4+7+8=25 Hence we will select only those five numbers whose sum is 18 or 9. But no five numbers will make the sum 9. So we choose the five numbers which make their sum 18. Since the sum of all the seven digits is 25 so we exclude those two digits out of given seven digits whose sum is 7.

Thus we exclude either O, 7 or 3, 4 as both these have sum 7. Hence we have the following two sets

$\text{ (a) }1,2,3,4,8,(0,7\text{ excluded) }\therefore \text{ Sum }=18$

They can be arranged in 5! = 120 ways 

(b) $0,1,2,7,8,(3,4$ excluded)  $\therefore \mathrm{Sum}=18$

$\therefore$ They can be arranged in 5! - 4! (0, in beginning}= 120 - 24 = 96 ways Thus the total number of numbers divisible by 9' is

$\Rightarrow$ 120 + 96 = 216