QJEEM20Ma4562: JEE Main 2020 Mathematics - Class 11 JEEM20Ma4562 Solution: The total number of 3-digit numbers, whose sum of digits is 10, …

Let xyz is 3 digits number.

Given that sum of digits = 10

$\therefore$ x + y + z = 10 ......(1)

Also x can't be 0 as if x = 0 then it will become 2 digits number.

So, x $\ge$ 1, y $\ge$ 0, z $\ge$ 0

As x $\ge$ 1

$\Rightarrow$ x $-$ 1 $\ge$ 0

Let x $-$ 1 = t

$\therefore$ t $\ge$ 0

From equation (1)

(x $-$ 1) + y + z = 9

$\Rightarrow$ t + y + z = 9

Now this problem becomes, distributing 9 things among 3 people t, y, z.

Number of ways we can do that

= ${}^{9 + 3 - 1}{C_{3 - 1}} = {}^{11}{C_2} = 55$

Now when 3 digit number is 900 then t = 9, y = 0, z = 0.

And when t = 9, then

x $-$ 1 = 9

$\Rightarrow$ x = 10

But we can't take x = 10 in a 3 digits number. So, we have to remove this case.

$\therefore$ Total number of 3 digit numbers = 55 $-$ 1 = 54.