Q157 / 270 · NDA 2026 (Phase 1) · Mathematics
A set $S$ contains $(2n+1)$ elements. If the number of subsets of $S$ which contain at most $n$ elements is $1024$, then what is the value of $n$ ?
- A$10$
- B$8$
- C$6$
- Correct answerD$5$
Solution
The number of elements in the set $S$ is $2n+1$.
The number of subsets of $S$ containing at most $n$ elements is given by the sum:
$\sum_{k=0}^{n} {}^{2n+1}C_{k}$
We know the sum of all binomial coefficients is:
$\sum_{k=0}^{2n+1} {}^{2n+1}C_{k} = 2^{2n+1}$
Using the symmetry property of binomial coefficients, $^{n}C_{r} = ^{n}C_{n-r}$, we have:
$\sum_{k=0}^{n} {}^{2n+1}C_{k} = \sum_{k=n+1}^{2n+1} {}^{2n+1}C_{k}$
Therefore, the sum of the first half of the coefficients is exactly half of the total sum:
$\sum_{k=0}^{n} {}^{2n+1}C_{k} = \dfrac{2^{2n+1}}{2} = 2^{2n}$
Given that the number of such subsets is $1024$, we can write:
$2^{2n} = 1024$
Since $1024 = 2^{10}$, we get:
$2n = 10$
$n = 5$
Answer: $5$
The number of subsets of $S$ containing at most $n$ elements is given by the sum:
$\sum_{k=0}^{n} {}^{2n+1}C_{k}$
We know the sum of all binomial coefficients is:
$\sum_{k=0}^{2n+1} {}^{2n+1}C_{k} = 2^{2n+1}$
Using the symmetry property of binomial coefficients, $^{n}C_{r} = ^{n}C_{n-r}$, we have:
$\sum_{k=0}^{n} {}^{2n+1}C_{k} = \sum_{k=n+1}^{2n+1} {}^{2n+1}C_{k}$
Therefore, the sum of the first half of the coefficients is exactly half of the total sum:
$\sum_{k=0}^{n} {}^{2n+1}C_{k} = \dfrac{2^{2n+1}}{2} = 2^{2n}$
Given that the number of such subsets is $1024$, we can write:
$2^{2n} = 1024$
Since $1024 = 2^{10}$, we get:
$2n = 10$
$n = 5$
Answer: $5$
Hence, the correct answer is option: 'D'.