Combinatorix

knight247 wrote:How many integers are there between 2*10^5 and 9*10^5 so that the sum of their digits is 3?
(A)9
(B)8
(C)7
(D)6
(E)5

Answer is D. Solved this without using combinatorix. Combinatorix being my weakness i'm hoping someone can solve this using combinatorix and clearly explain the steps. Thanks

Forget combination theory - it would just confuse in this case. For these small answer choices, it's best to simply count.

We start with 2*10^5 = 200000. We already have a 2, so to reach the sum of digits of 3, we just need to add 1. That 1 can come replace one of the 5 zeroes:
200001
200010
200100
201000
210000

5 options.

Next, move to the 300000 - and stop there. Already we have a sum of digits of 3: 3+0+0+0+0+0. We can't add any other digit, or we get a sum of digits greater than 3.

The same can be said for any number beginning with 4, or 5, or 6 - they're already greater than the sum of digits of 3, so they're out. So there are 6 options altogether - five options beginning with 2, and one option beginning with 3.

Only these six possibilities are there.

200001
200010
200100
201000
210000
300000

Hence answer should be D

possible combinations are 2,1 and 3,0.

5 positions for 1 hence 2,1 = 5
1 position for 3 hence 3,0 = 1

thus 6 in total.