Permutations & Combinations
Permutation and Combination
star_batch_jee_advanced_2025
Grade 11
If the number of ways in which 8 people can be arranged in a line if $A$ and $B$ must be next to each other and $C$ must be somewhere behind $D$ is equal to $'m'$ then sum of all the digits of $m$ is equal to ______.
Step-by-Step Solution
Key Concept: Use the unit method for the adjacency constraint, then apply symmetry to handle the relative ordering constraint by dividing by 2.
First, treat $A$ and $B$ as a single unit since they must be adjacent. This gives us 7 units to arrange (the $AB$ unit plus 6 other people), which can be done in $7!$ ways. Within the $AB$ unit, $A$ and $B$ can be arranged in $2!$ ways, giving $7! \times 2! = 5040 \times 2 = 10080$ arrangements. Now apply the constraint that $C$ must be behind $D$: in any arrangement of the 8 people, $C$ and $D$ can be in $2!$ relative orders (either $D$ before $C$ or $C$ before $D$), and by symmetry, exactly half have $D$ before $C$. Therefore, $m = \frac{10080}{2} = 5040$. The sum of digits of $5040$ is $5 + 0 + 4 + 0 = 9$.
Correct Answer: 9