The number of integers between $$1$$ and $$1000000$$ that have the sum of the digits $$18$$, is

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Infinity Learn · Accepted Answer

Any number between $$1$$ and $$1000000$$ must be of less than seven digits. Therefore, it must be of the form $$a1$$ $$a2$$ $$a3$$ $$a4$$ $$a5$$ $$a6where$$ $$a1$$, $$a2$$, $$a3$$, $$a4$$, $$a5$$, $$a6$$ ∈ $${0$$, $$1$$, $$2$$, …, $$9}According$$ to question, sum of the digits $$=$$ $$18Thus$$, $$a1$$ $$+$$ $$a2$$ $$+$$ $$a3$$ $$+$$ $$a4$$ $$+$$ $$a5$$ $$+$$ $$a6$$ $$=$$ $$18where$$ $$0$$ ≤ ai ≤ $$9$$, i $$=$$ $$1$$, $$2$$, $$3$$, …, $$9$$.Required $$number=$$ coefficient of $$x18$$ in $$1+x+x2+$$…$$+x96=$$ coefficient of $$x18$$ in $$1$$−$$x101$$−$$x6=$$ coefficient of $$x18$$ in $$1$$−$$x106(1$$−$$x)$$−$$6=$$ coefficient of $$x18$$ in $$1$$−$$6C1x10(1$$−$$x)$$−$$6$$[leaving terms containing powers of x greater than $$18$$]$$=$$ coefficient of $$x18$$ in (1$$−$$x)−$$6$$−$$6C1$$. coefficient of $$x8$$ in (1$$−$$x)−$$6$$ $$=6+18$$−$$1C18$$−$$6$$⋅$$6+8$$−$$1C18=23C5$$−$$6$$⋅$$13C8=23$$⋅$$22$$⋅$$21$$⋅$$20$$⋅$$19120$$−$$6$$⋅$$13$$⋅$$12$$⋅$$11$$⋅$$10$$⋅$$9120=33649$$−$$7722=25927$$

The number of integers between 1 and 1000000 that have the sum of the digits 18, is

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