Question

Let $A$ and $B$ be two non-empty sets with $n(A)=4$ and $n(B)=5$. If a mapping is selected at random from the set of all mappings from $A$ to $B$, then the probability of getting a many-one mapping is

• A. $\dfrac{29}{125}$
• B. $\dfrac{24}{125}$
• C. $\dfrac{96}{125}$
• D. $\dfrac{101}{125}$

Hint:

Many-one mappings = Total mappings $-$ One-one mappings.

Answer 1

The Correct Option is D

Step 1: Count all mappings.
Each of the $4$ elements of $A$ can go to any of the $5$ elements of $B$. So the total number of mappings is $5^4=625.$ This is the denominator.
Step 2: Understand many-one.
A mapping is many-one if at least two elements of $A$ share the same image. It is easier to count the opposite, the one-one mappings, and subtract.
Step 3: Count one-one mappings.
A one-one map sends the $4$ elements to $4$ different images chosen from $5$: $5\times4\times3\times2=120.$
Step 4: Get many-one mappings.
Many-one $=$ total $-$ one-one $=625-120=505.$
Step 5: Form the probability.
$P=\dfrac{505}{625}.$
Step 6: Simplify.
Divide top and bottom by $5$: $\dfrac{101}{125}.$ \[ \boxed{\dfrac{101}{125}} \]