What is the number of $6$ digit positive integers whose sum of the digits is at least $52$?

Question

What is the number of $6$ digit positive integers whose sum of the digits is at least $52$?

My kind of approach was:

I thought to use Multinomial Theorem concept here.

So my primary aim was to find the coefficient of $x^{52}$in {followed by $x^{53}$ and $x^{54}$[$54$ will be highest because $6\cdot 9=54$]} $$(x^{1}+x^{2}+x^{3}+\ldots+x^{9})^{6};$$ $=\text{finding coefficient of } x^{52 } \text{ from } x^{6}\cdot (1-x^{9})^{6}\cdot (1-x)^{-6}$ $=\text{this is equivalent to finding coefficient of } x^{46 } \text{ from } (1-x^{9})^{6}\cdot (1-x)^{-6}$

$=(1-x^{9}+x^{18}-x^{27}+x^{36}-x^{45})\cdot (1-x)^{-6}$[rest term will be a waste]

This leads to: $\binom{51}{46}-\binom{42}{37}+\binom{33}{28}-\binom{24}{19}+\binom{15}{10}-\binom{6}{1}$

But it is a wrong answer. How to use Multinomial correctly?

Answer 1

Note that $6\cdot 9=54$ so having 52 is hard - it requires very high digits.

You can think of it as starting with 999999 (sum=54) and then removing 2. You can either do it in two places (turning two 9 into 8) or in one place (turn one 9 into 7). The number of options is $${6 \choose 2}+6$$ To get 53 you need to remove only 1 (6 options) and to get 54 you have one the above- mentioned options.

To conclude, the number of 6 digit integers with sum of at least 52 is $${6 \choose 2}+6+6+1$$

Answer 2

You should have: $$[x^{52}](1+x+x^2+\cdots +x^9)^6=[x^{52}]\left(\frac{1-x^{10}}{1-x}\right)^6=\\ [x^{52}](1-x^{10})^6\cdot (1-x)^{-6}=\\ [x^{52}]\sum_{i=0}^6 {6\choose i}(-x^{10})^i\cdot \sum_{j=0}^{\infty}{-6\choose j}(-x)^j=\\ [x^{52}]\sum_{i=0}^6 {6\choose i}(-x^{10})^i\cdot \sum_{j=0}^{\infty}{6+j-1\choose j}x^j=\\ \small{57\choose 52}-{6\choose 1}{47\choose 42}+{6\choose 2}{37\choose 32}-{6\choose 3}{27\choose 22}+{6\choose 4}{17\choose 12}-{6\choose 5}{7\choose 2}=21$$ Wolfram answer.

Indeed, to have the sum of $52$, the digits should be $799999$ or $889999$: $${6\choose 1}+\frac{6!}{2!4!}=21.$$