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I saw a problem of this kind sometime ago. It was solved using the coefficient of something in binomial theorem. I'm not sure. I am unable solve this. Thanks in advance.

user585278
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3 Answers3

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Imagine that we don't have our usual digits -- we still want to use base-10 notation, but have to write each digit by tally marks, with some spacer such as a colon between different powers of ten. So 123 would be written I:II:III and 307 would be written III::IIIIIII.

Now you're looking for 3-digit number -- so in our impoverished notation it contains two colons, and it doesn't start with a colon because then it would really be a 2-digit number. And the sum of the digits is eleven, so there are eleven Is.

In other words, what we're looking for is I followed by some combination of two colons and ten further Is. There are $\binom{12}{2}$ possibilities for this, but a few of them don't count, namely the ones where one of the digits is ten or more. But those are easy to enumerate: They are exactly XI:: and X:I: and X::I and I:X: and I::X.

So subtract $5$ from $\binom{12}{2}$.

hmakholm left over Monica
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Try to solve the following question:

How many two digit integers are there such that the sum of the digits is $k$, where $2\leq k\leq 10$? (Hint: after you choose the first digit, how many choices do you have for the second?)

Next, using the previous idea, how many 3 digit integers meeting your requirement are there that start with $1$? How about $2$? Can you finish from here?

TomGrubb
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Think of a three digit number as picking a number one through nine for the first digit, then any number zero through next two. So there are $9\cdot 10\cdot 10$ three digit numbers, and the number of these whose coefficients add to $11$ is the coefficient of $x^{11}$ in the expression

$$(x+x^2+\cdots+x^9)(1+x+\cdots+x^9)(1+x+\cdots+x^9).$$

To find this coefficient, you can ask a computer, or use the multinomial theorem. I got the answer 61.

pancini
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