How many numbers between $0$ and $1,000,000$ have exactly one digit equal to $9$ and the sum of digits equal $13$?
I am not sure how to start this problem to get this answer. Could someone help me with this problem?
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If one digit is $9$ and the sum is $13$, the other digits add up to $4$. Not too many possibilities to consider.
Robert Israel
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What you want to count is the number of sequences of digits of length $6$ in which the sum of digits is $13$ and there is exactly one $9$.
There are $6$ places where we can put the $9$. Once you have placed the $9$ how many options are there for the other positions? We just have to make sure that the sum of the numbers in the five positions is $4$.
Since $4$ is less than $10$ we can solve it easily with stars and bars, there are $5$ digit positions and they add up to $4$, so it is $\binom{4+5-1}{4}$. So the final answer is $6\binom{5+4-1}{4}=6\binom{8}{4}=420$
Asinomás
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If I understand your argument, you are choosing one of the six digits to be $9$, then calculating the number of ways that the sum of the other five digits can add to $6$. Your first paragraph does not make this clear. – N. F. Taussig Feb 01 '16 at 18:01
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Not quite. Since there are six digits, there are six places we can put the $9$, so your original answer was correct. You need to fix the second paragraph so that the answer you had makes sense. – N. F. Taussig Feb 01 '16 at 18:12
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{1, 1, 1, 1}, {1, 1, 2}, {2, 2}, {1, 3}, {4}– karakfa Feb 01 '16 at 16:34