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

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