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

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 this can be done is

Infinity Learn · Accepted Answer

We have $$0$$ $$+$$ $$1$$ $$+$$ $$2$$ $$+$$ $$3$$ … $$+$$ $$8$$ $$+$$ $$9$$ $$=$$ $$45To$$ 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.$${0$$ can not be at extreme left.$$}Hence$$, there are $$8$$!$$+4(8$$!−$$7$$!$$)=(36)(7$$!$$)$$ numbers in the  desired category.

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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