Rosen
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in Combinatory closed by
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closed with the note: got the answer and to avoid unnecessary discussion.

How many positive integers less than 1,000,000 have the
sum of their digits equal to 19?

Doubt 1: I understood till C(24,5). Now while subtracting unfavorable cases, for the sum not to exceed 19 one or more digits cannot exceed 9 coz 10+10= >19. But each digit can only have values from 0-9. how could a digit take a value 10?

My approach : It is similar to selection with repetition allowed. We have to select 19 balls from 6 boxes. Now to violate the condition, not more than 9 balls should be removed from more than one box, otherwise it will be 20. So suppose I take out 10 balls from box 1, still 9 more balls are remaining. It is same as selecting 9 balls from 6 boxes with repetition allowed. Then how it is unfavorable according to solution given?

plz correct me.

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4 Comments

surely it's my mistake...
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@Shaik Masthan

It will be  6C1 Because we have to just select (choose) the one Objects among the 6 objects.We don't need any type of arrangement here.

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agreed brother...
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