Let A and B be finite sets and P_A and P_B re | vedclass

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

Question

(D) Let $n(A) = m$ and $n(B) = n$. Given that $P_B$ has $112$ elements more than $P_A$,we have $n(P_B) - n(P_A) = 112$. Since $n(P_A) = 2^m$ and $n(P_

Vedclass · Accepted Answer

$840$

Vedclass · Answer

(D) Let $n(A) = m$ and $n(B) = n$. Given that $P_B$ has $112$ elements more than $P_A$,we have $n(P_B) - n(P_A) = 112$. Since $n(P_A) = 2^m$ and $n(P_B) = 2^n$,we get $2^n - 2^m = 112$. Factoring out $2^m$,we have $2^m(2^{n-m} - 1) = 112$. We can write $112$ as $16 	imes 7 = 2^4 	imes (8 - 1) = 2^4(2^3 - 1)$. Comparing both sides,we get $m = 4$ and $n - m = 3$,which implies $n = 7$. The number of injective functions from $A$ to $B$ is given by the formula $^n P_m$. Substituting the values,we get $^7 P_4 = \frac{7!}{(7-4)!} = \frac{7 	imes 6 	imes 5 	imes 4 	imes 3!}{3!} = 7 	imes 6 	imes 5 	imes 4 = 840$.

Let $A$ and $B$ be finite sets and $P_A$ and $P_B$ respectively denote their power sets. If $P_B$ has $112$ elements more than those in $P_A$,then the number of injective functions from $A$ to $B$ is

Similar Questions

If a set $A$ has $5$ elements,then the number of ways of selecting two subsets $P$ and $Q$ from $A$ such that $P$ and $Q$ are mutually disjoint,is

Find the simplest form of the following switching circuit.

Let $S = \{1, 2, 3, 4\}$. The total number of unordered pairs of disjoint subsets of $S$ is equal to

If $P(x) = x^5 + ax^4 + bx^3 + cx^2 + dx + e$ is a polynomial such that $P(0) = 1, P(1) = 2, P(2) = 5, P(3) = 10$ and $P(4) = 17$,then $P(5) =$

If $U$ is the universal set and $A \cup B \cup C = U$,then ${(A - B) \cup (B - C) \cup (C - A)}'$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module