A seven digit number divisible by 9 is to be formed by using 7 out of number $$\left\{ {1,2,3,4,5,6,7,8,9} \right\}.$$     The number of ways in which this can be done is

Solution :
Sum of $$7$$ digits $$= a$$  multiple of $$9$$ Since sum of number $$1, 2, 3, . . . . . , 8, 9$$    is $$45\,$$ (Since a number is divisible by $$9$$ if sum of its digits is divisible by $$9.$$ ) So, two left number should also have sum as $$9.$$ The pairs to be left are $$\left( {1,8} \right),\left( {2,7} \right),\left( {3,6} \right),\left( {4,5} \right).$$      With each pair left, number of $$7$$ digit numbers $$= 7!.$$  So, with all $$4$$ pairs total seven digits number $$= 4 × 7!$$