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  4. 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 this can be done is

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Q. 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 this can be done is

Permutations and Combinations

Solution:

We have $0+1+2+3 \ldots+8+9=45$
To obtain an eight digit number exactly divisible by 9 , we must not use either $(0,9)$ or $(1,8)$ or $(2,7)$ or $(3,6)$ or $(4,5)$. [Sum of the remaining eight digits is 36 which is exactly divisible by 9 .]
When, we do not use $(0,9)$, then the number of required 8 digit numbers is $8 !$.
When one of $(1,8)$ or $(2,7)$ or $(3,6)$ or $(4,5)$ is not used, the remaining digits can be arranged in 8 !- 7 ! ways. $\{0$ cannot be at extreme left.
Hence, there are $8 !+4(8 !-7 !)=(36)(7 !)$ numbers in the desired category.