Permutations & Combinations
Permutations & Combinations
star_batch_jee_advanced_2025
Grade None
Question:
You are given an unlimited supply of each of the digits $1, 2, 3$ or $4$. Using only these four digits, you construct $n$ digit numbers. Such $n$ digit numbers will be called LEGITIMATE if it contains the digit $1$ either an even number times or not at all. Number of $n$ digit legitimate numbers are:
$2^n + 1$
$2^{n+1} + 2$
$2^{n+2} + 4$
$2^{n-1}(2^n + 1)$
Step-by-Step Solution
Key Concept: Separating the fraction into partial binomial sums allows telescoping and simplification.
The sum $\sum_{r=1}^{n} \frac{r(r+1)}{2r}C_r^2$ is simplified by expressing each term using binomial coefficient identities. The expression separates into three parts involving $^nC_r^2$, $^{n-1}C_r$ and related terms. After algebraic manipulation and telescoping, the sum evaluates to $\frac{1}{2}(2^{2n-1}C_{n-1} + ^{2n}C_n - 1)$.
Correct Answer: 4