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Question

Mathematics Question on Relations

Let AA and BB be finite sets and PAP_{A} and PBP_{B} respectively denote their power sets. If PBP_{B} has 112112 elements more than those in PAP_{A^{\prime}} then the number of functions from AA to BB which are injective is

A

224

B

56

C

120

D

840

Answer

840

Explanation

Solution

Let n(A)=mn(A)=m and n(B)=nn(B)=n. Then, according to the question, n(P(B))n(P(A))=112n(P(B))-n(P(A)) =112 2n2m=112\Rightarrow 2^{n}-2^{m} =112 2m(2nm1)=16×7\Rightarrow 2^{m}\left(2^{n-m}-1\right) =16 \times 7 2m(2nm1)=24(231)\Rightarrow 2^{m}\left(2^{n-m}-1\right) =2^{4}\left(2^{3}-1\right) m=4\therefore m=4 and nm=3 n-m =3 m=4,n=7\Rightarrow m=4, n=7 \therefore Number of injective functions from AA to BB =n(B)Pn(A)=7P4=7!3!=840={ }^{n(B)} P_{n(A)}={ }^{7} P_{4}=\frac{7 !}{3 !}=840