Question

How many positive integers less than 1,000,000 have the sum of their digits equal to 19?

Short answer

There are 30,492 positive integers less than 1,000,000 have the sum of their digits equal to 19

Step-by-step solution

Step 1: Use the formula for integer

Formula for integer

\(C\left( {n + r - 1,r} \right) = \frac{{(n + r - 1)!}}{{n!r!}}\)

n is number of distinguishable boxes

r is number of distinguishable objects

Step 2: Solution of number of 19 distinguishable objects

Let’s, applied n and r values of inequality in integer formula

A number less than 1,000,000 will have at most 6 digits.

\({x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6} = 19\)

Ways to select 19 distinguishable objects from 6 distinguishable boxes

\(\begin{array}{l}C\left( {n + r - 1,r} \right) = \frac{{(n + r - 1)!}}{{n!r!}}\\n = 6\\r = 19\\C\left( {n + r - 1,r} \right) = C\left( {6 + 19 - 1,19} \right)\\ = C\left( {24,19} \right)\\ = \frac{{24!}}{{6!19!}}\\ = 42,504\end{array}\)

Step 3: Solution of number of 9 distinguishable objects

Let’s, applied n and r values of inequality in integer formula

A number less than 1,000,000 will have at most 6 digits.

\({x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6} = 9\)

Ways to select 9 distinguishable objects from 6 distinguishable boxes

\(\begin{array}{l}C\left( {n + r - 1,r} \right) = \frac{{(n + r - 1)!}}{{n!r!}}\\n = 6\\r = 9\\C\left( {n + r - 1,r} \right) = C\left( {6 + 9 - 1,9} \right)\\ = C\left( {14,9} \right)\\ = \frac{{14!}}{{6!9!}}\\ = 2002\end{array}\)

Step 4: Solution of number of sum greater than 19

We can only have 1 Xi because 2Xis will make the sum greater than 19.

Since Xi can take 6 values we multiply 2002 by 6 i.e., 12,012.

So, 42,504 solutions are there without restrictions, but with the restriction

we have only

\(42,504 - 12,012{\rm{ }} = {\rm{ }}30,492\)