The total number of 3-digit numbers, whose sum of digits is 10, is __________.

Let xyz is 3 digits number.<br><br>Given that sum of digits = 10<br><br>$\therefore$ x + y + z = 10 ......(1)<br><br>Also x can't be 0 as if x = 0 then it will become 2 digits number.<br><br>So, x $\ge$ 1, y $\ge$ 0, z $\ge$ 0<br><br>As x $\ge$ 1<br><br>$\Rightarrow$ x $-$ 1 $\ge$ 0<br><br>Let x $-$ 1 = t<br><br>$\therefore$ t $\ge$ 0<br><br>From equation (1)<br><br>(x $-$ 1) + y + z = 9<br><br>$\Rightarrow$ t + y + z = 9<br><br>Now this problem becomes, distributing 9 things among 3 people t, y, z.<br><br>Number of ways we can do that<br><br>= ${}^{9 + 3 - 1}{C_{3 - 1}} = {}^{11}{C_2} = 55$<br><br>Now when 3 digit number is 900 then t = 9, y = 0, z = 0.<br><br>And when t = 9, then<br><br>x $-$ 1 = 9<br><br>$\Rightarrow$ x = 10<br><br>But we can't take x = 10 in a 3 digits number. So, we have to remove this case.<br><br>$\therefore$ Total number of 3 digit numbers = 55 $-$ 1 = 54.