Q.If the set $A$ contains 5 elements and the set $B$ contains 6 elements, then the number of one-one and onto mappings from $A$ to $B$ is
(A) 720
(B) 120
(C) 0
(D) none of these

The key idea is that a one-one and onto mapping (a bijection) requires the two sets to have the same number of elements.

• Set $A$ has $5$ elements, set $B$ has $6$ elements.
• A function from $A$ to $B$ that is both injective (one-one) and surjective (onto) must pair each element of $A$ with a distinct element of $B$, and every element of $B$ must be used. …

Why counting functions needs the right lens

When we talk about mappings from $A$ to $B$, we are assigning each element of $A$ to some element of $B$. A one‑one (injective) function means no two elements of $A$ map to the same element of $B$. An onto (surjective) function means every element of $B$ is the image of at least one element of $A$. A function that is both one‑one and onto is called a bijection.

The key constraint for a bijection is a simple size condition: the two sets must have the same cardinality. Why? Because a one‑one function from $A$ to $B$ can use at most $|A|$ distinct elements of $B$, and an onto function must use all $|B|$ elements. For both to hold simultaneously, we need $|A| = |B|$.

Here, $|A| = 5$ and $|B| = 6$. They are not equal. So no bijection exists — the count is zero.

Let’s walk through the reasoning step by step.

1. Define the problem precisely

   We need the number of functions $f: A \to B$ that are both injective (one‑one) and surjective (onto). Such functions are called bijections.

2. Check the necessary condition

   For a bijection to exist, the domain and codomain must have the same number of elements.

   • If $f$ is one‑one, then $|A| \le |B|$ (pigeonhole principle).
   • If $f$ is onto, then $|A| \ge |B|$ (every element of $B$ must be hit).
   • Combining: $|A| = |B|$.

   Here $5 \neq 6$, so the condition fails. …

Method: Counting one-one and onto maps, and the size condition

Use this when asked for the number of injective, surjective, or bijective mappings between two finite sets.

Steps

Step 1: Check the feasibility (size) condition FIRST.

A one-one map needs $|A|\le|B|$; an onto map needs $|A|\ge|B|$. A map that is both (a bijection) needs $|A|=|B|$. Testing this before any calculation often settles the problem instantly.

Step 2: If the size condition fails, the count is $0$.

Here $|A|=5$ and $|B|=6$ are unequal, so no map can be both one-one and onto — the answer is $0$, with no permutation calculation needed.

Step 3: When it is feasible, count with the right formula. …

Common Mistakes

Mistake 1: Computing $^{6}P_{5}=720$ and reporting it as the answer.

Why it's wrong: $720$ counts one-one maps, but such maps cannot also be onto when $|A|=5<|B|=6$ — five arrows can never cover six elements. Correct approach: check the size condition first; the count of one-one AND onto maps is $0$.

Mistake 2: Forgetting that a bijection forces $|A|=|B|$.

Why it's wrong: one-one needs $|A|\le|B|$ and onto needs $|A|\ge|B|$, so together they demand equal sizes. Correct approach: since $5\neq 6$, conclude immediately that no such mapping exists. …