QJEEM23Maee99: JEE Main 2023 Mathematics - Class 11 JEEM23Maee99 Solution: Number of 4-digit numbers (the repetition of digits is allowed) …

We have to make 4 digit numbers using the digits, 1, 2, 3 and 5.

The unit digit of the 4 digit number will be 5.

Now, the sum (x + y + z) should be of the (3p + 1).

Therefore, the possible cases are

(x, y, z) = (1, 1, 5), (1, 1, 2), (2, 2, 3), (2, 3, 5), (3, 3, 1) and (5, 5, 3).

So, total arrangements are

For $(1,1,5) \rightarrow \frac{3 !}{2 !}=3 ;$

For $(1,1,2) \rightarrow \frac{3 !}{2 !}=3 ;$

For $(2,2,3) \rightarrow \frac{3 !}{2 !}=3 ;$

For $(2,3,5) \rightarrow 3 !=6 ;$

For $(3,3,1) \rightarrow \frac{3 !}{2 !}=3 ;$

For $(5,5,3) \rightarrow \frac{3 !}{2 !}=3 ;$

So, total number of arrangements $=3+3+3+6+3+3=$ 21