- What is the total number of 3-digit numbers, whose sum of digits is 10?

- Hint: Let the three digits be \[x,y,z\]. Clearly, \[1 \le x \le 9,0 \le y \le 9,0 \le z \le 9\]. Then use the rule of expansion of a 3-digit number and make an equation following the condition “sum of the three digits is 10”. After that put the different values of \[x,y,z\] and find out the required answer.

Formula used: The expansion of a 3 digit number is \[100x + 10y + z\], where \[x,y,z\] are the digits of the number at hundred’s place, ten’s place and unit’s place respectively.

Complete step-by-step solution: Let \[x,y,z\] are the digits of a 3 digit number at hundred’s place, ten’s place and unit’s place respectively.
Then the number in the form of expansion is \[100x + 10y + z\].
It is given that the sum of the digits is \[10\].
So, \[x + y + z = 10......\left( i \right)\], where \[1 \le x \le 9,0 \le y \le 9,0 \le z \le 9\]
If \[x = 1\], then \[y + z = 9 \Rightarrow z = 9 - y\]
Putting \[y = 0\], we get \[z = 9\]
Putting \[y = 1\], we get \[z = 8\]
Putting \[y = 2\], we get \[z = 7\]
Putting \[y = 3\], we get \[z = 6\]
Putting \[y = 4\], we get \[z = 5\]
Putting \[y = 5\], we get \[z = 4\]
Putting \[y = 6\], we get \[z = 3\]
Putting \[y = 7\], we get \[z = 2\]
Putting \[y = 8\], we get \[z = 1\]
Putting \[y = 9\], we get \[z = 0\]
So, the numbers are \[109,118,127,136,145,154,163,172,181,190\]
\[\therefore \] From \[100\] to \[199\], there are total \[10\] such numbers.
Similarly, putting \[x = 2\] in equation \[(i)\], we get \[y + z = 8\]
Putting \[y = 0,1,2,3,4,5,6,7,8\] respectively, we get \[z = 8,7,6,5,4,3,2,1,0\]
We can’t put \[y = 9\] because putting \[y = 9\], we get \[z = - 1\], which is not possible.
So, the numbers are \[208,217,226,235,244,253,262,271,280\]
\[\therefore \] From \[200\] to \[299\], there are total \[9\] such numbers.
Similarly, putting \[x = 3\], we get the numbers \[307,316,325,334,343,352,361,370\]
\[\therefore \]From \[300\] to \[399\], there are total \[8\] such numbers.
Similarly, putting \[x = 4\], we get the numbers \[406,415,424,433,442,451,460\]
\[\therefore \]From \[400\] to \[499\], there are total \[7\] such numbers.
Similarly, putting \[x = 5\], we get the numbers \[505,514,523,532,541,550\]
\[\therefore \]From \[500\] to \[599\], there are total \[6\] such numbers.
Similarly, putting \[x = 6\], we get the numbers \[604,613,622,631,640\]
\[\therefore \]From \[600\] to \[699\], there are total \[5\] such numbers.
Similarly, putting \[x = 7\], we get the numbers \[703,712,721,730\]
\[\therefore \]From \[700\] to \[799\], there are total \[4\] such numbers.
Similarly, putting \[x = 8\], we get the numbers \[802,811,820\]
\[\therefore \]From \[800\] to \[899\], there are total \[3\] such numbers.
Similarly, putting \[x = 9\], we get the numbers \[901,910\]
\[\therefore \]From \[900\] to \[999\], there are total \[2\] such numbers.
Hence, we get total \[10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 = 54\] such numbers.

Note: Many student mistake while putting the values of \[x,y,z\] and using the condition \[1 \le x \le 9,0 \le y \le 9,0 \le z \le 9\] properly. Since, the digits must be of 3 digits, so the digit of hundred’s place should not be equal to zero.