Amit M Agarwal Solutions for Chapter: Binomial Theorem, Exercise 3: Work Book Exercise 8.3

There are two bags each of which contains $n$ balls. A man has to select an equal number of balls from both the bags. The number of ways in which a man can choose at least one ball from each bag is

$P$ is a set containing $n$ elements. A subset $A$ of $P$ is chosen and the set $P$ is reconstructed by replacing the elements of $A.$ A subset $B$ of $P$ is chosen again. The number of ways of choosing $A$ and $B$ such that $A$ and $B$ have no common elements is

$P$ is a set containing $n$ elements. A subset $A$ of $P$ is chosen and the set $P$ is reconstructed by replacing the elements of $A.$ A subset $B$ of $P$ is chosen again. The number of ways of choosing $A$ and $B$ such that $A=B$, is

$P$ is a set containing $n$ elements. A subset $A$ of $P$ is chosen and the set $P$ is reconstructed by replacing the elements of $A.$ A subset $B$ of $P$ is chosen again. The number of ways of choosing $A$ and $B$ such that $B$ is a subset of $A$, is

$P$ is a set containing $n$ elements. A subset $A$ of $P$ is chosen and the set $P$ is reconstructed by replacing the elements of $A$. A subset $\mathrm{\Sigma }$ of $P$ is chosen again. The number of ways of choosing $A$ and $B$ such that $B$ contains just one element more than $A$, is

$P$ is a set containing $n$ elements. A subset $A$ of $P$ is chosen and the set $P$ is reconstructed by replacing the elements of $A$. A subset $B$ of $P$ is chosen again. The number of ways of choosing $A$ and $B$ such that $A$ and $B$ have the equal number of elements, is

If $n>3$, then $xy{C}_0-(x-1)(y-1)C_1+(x-2)(y-2)C_2-(x-3)(y-3)C_3+\dots +(-1{)}^n(x-n\left)\right(y-n)C_n$ equals

If $n>3,$ then $xyz{C}_0-(x-1)(y-1)(z-1)C_1+(x-2)(y-2)(z-2)C_2-(x-3)(y-3)(z-3)C_3+\dots +(-1{)}^n(x-n\left)\right(y-n\left)\right(z-n)C_n$ equals