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 $Q$ i

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

The sum of all $4$-digit numbers that can be formed using the digits $2, 3, 4, 5, 6$ without repetition is:

If ${ }^{n} P_{r}={ }^{n} P_{r+1}$ and ${ }^{n} C_{r}={ }^{n} C_{r-1}$,then the value of $r$ is equal to:

If a number is chosen at random from the four-digit numbers formed by using the digits $0, 1, 2, 3, 4, 6$ without repetition,then the probability that it is divisible by $4$ is

The number of positive integral solutions of the equation $xyz = 3000$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module