Question

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 $P_{A^{\prime}}$ then the number of functions from $A$ to $B$ which are injective is

• A. 224
• B. 56
• C. 120
• D. 840

Answer 1

The Correct Option is D

Let $n(A)=m$ and $n(B)=n$. Then, according to the question, $ n(P(B))-n(P(A)) =112 $ $\Rightarrow 2^{n}-2^{m} =112 $ $\Rightarrow 2^{m}\left(2^{n-m}-1\right) =16 \times 7$ $\Rightarrow 2^{m}\left(2^{n-m}-1\right) =2^{4}\left(2^{3}-1\right) $ $\therefore m=4 $ and $ n-m =3$ $\Rightarrow m=4, n=7$ $\therefore$ Number of injective functions from $A$ to $B$ $={ }^{n(B)} P_{n(A)}={ }^{7} P_{4}=\frac{7 !}{3 !}=840$