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The answer is D-36(7!).. :-)
Explanation: First the sum of the 10 digits (0,1,...,9) is 45,which is divisible by 9.So we have to wisely remove 2 digits from here so that the sum remains divisible by 9. now 5 combination of such 2 digits are possible ie. (0,9),(1,8),(2,7),(3,6) and (4,5).for (0,9) combination the permutation of rest 8 digits are 8!.For each of rest 4 combinations the possible combination 8 digits numbers is=(8!-7!)as we removing the case where 0 is the leading digit..
so total possible combinations is:
8!+4*(8!-7!)
=(8*7!)+4*7!(8-1)
=(8*7!)+28*7!
=(8+28)*7!
=36(7!)(required ans..)
Hope it was helpful......
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The answer is D-36(7!).. :-) Explanation: First the sum of the 10 digits (0,1,...,9) is 45,which is divisible by 9.So we have to wisely remove 2 digits from here so that the sum remains divisible by 9. now 5 combination of such 2 digits are possible ie. (0,9),(1,8),(2,7),(3,6) and (4,5).for (0,9) combination the permutation of rest 8 digits are 8!.For each of rest 4 combinations the possible combination 8 digits numbers is=(8!-7!)as we removing the case where 0 is the leading digit.. so total possible combinations is: 8!+4*(8!-7!) =(8*7!)+4*7!(8-1) =(8*7!)+28*7! =(8+28)*7! =36(7!)(required ans..) Hope it was helpful......