JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Permutations and Combinations

The number of seven digit integers with sum of the digits equal to 10 and formed by using the digits 1,2 and 3 only is:

Answer: (A)

Solution

We need to find non-negative integer solutions to

x_{1} + 2 x_{2} + 3 x_{3} = 10

with

x_{1} + x_{2} + x_{3} = 7

, where

x_{i}

represents the count of digit

i

. From the second equation,

x_{1} = 7 - x_{2} - x_{3}

. Substituting into the first:

(7 - x_{2} - x_{3}) + 2 x_{2} + 3 x_{3} = 10 \Rightarrow x_{2} + 2 x_{3} = 3

. Valid solutions:

(x_{2}, x_{3}) \in {(3, 0), (1, 1)}

, giving

(x_{1}, x_{2}, x_{3}) \in {(4, 3, 0), (5, 1, 1)}

. The number of arrangements:

\frac{7 !}{4 ! 3 !} + \frac{7 !}{5 ! 1 ! 1 !} = 35 + 42 = 77

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Single correct option