Let A be a set containing n elements. A subse | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $|P| = k$. Then $Q$ must have $k+1$ elements.For a fixed $k$,the number of ways to choose $P$ is $^{n}C_k$.The number of ways to choose $Q$ is

Vedclass · Accepted Answer

$^{2n}C_{n-1}$

Vedclass · Answer

(A) Let $|P| = k$. Then $Q$ must have $k+1$ elements.For a fixed $k$,the number of ways to choose $P$ is $^{n}C_k$.The number of ways to choose $Q$ is $^{n}C_{k+1}$.Since $P$ and $Q$ are chosen independently,the total number of ways is the sum over all possible values of $k$ from $0$ to $n-1$:$\sum_{k=0}^{n-1} (^{n}C_k \cdot ^{n}C_{k+1})$Using the identity $^{n}C_k = ^{n}C_{n-k}$,we have:$\sum_{k=0}^{n-1} (^{n}C_{n-k} \cdot ^{n}C_{k+1})$This is the coefficient of $x^{n+1}$ in the expansion of $(1+x)^n \cdot (1+x)^n = (1+x)^{2n}$,which is $^{2n}C_{n+1}$.Note that $^{2n}C_{n+1} = ^{2n}C_{2n-(n+1)} = ^{2n}C_{n-1}$.

Let $A$ be a set containing $n$ elements. $A$ subset $P$ of $A$ is chosen,and the set $A$ is reconstructed by replacing the elements of $P$. $A$ subset $Q$ of $A$ is chosen again. The number of ways of choosing $P$ and $Q$ such that $Q$ contains just one element more than $P$ is

Similar Questions

An examination has $3$ multiple-choice questions,and each question has $4$ options. If a student passes only if they answer all questions correctly,in how many ways can they fail?

If $f(n) = n! (31-n)!$,where $n \in \{0, 1, 2, \ldots, 31\}$,then the minimum value of $f(n)$ is

The number of all four-digit numbers which do not have four distinct digits is

The number of ways in which a committee of $7$ members can be formed from $6$ teachers,$5$ fathers,and $4$ students in such a way that at least one from each group is included and teachers form the majority among them,is

$^{80}C_{40}$ is not divisible by -

Vedclass Test Series

Exam Paper Generator

Online Exam Module