AP EAMCET · Maths · Functions

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 functions from $A$ to $B$ which are injective is

A $224$
B $56$
C $120$
D $840$

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Answer & Solution

Correct Answer

(D) $840$

Step-by-step Solution

Detailed explanation

Consider NA=m and NB=n PB has 112 elements more than PA NPB=112+NPA ⇒NPB-NPA=112 ⇒2n-2m=16×7 Rewrite the above equation. 2m2n-m-1=16×7 ⇒2m2n-m-1=2423-1 From the above equation m=4 ⇒n-m=3 ⇒n-4=3 ⇒n=7 The number of injective…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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

Step-by-step Solution

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 functions from $A$ to $B$ which are injective is