Question

If a set \(A\) has \(m\)-elements and the set \(B\) has \(n\)-elements, then the number of injections from \(A\) to \(B\) is

• A. \({}^{n}C_{m}\), if \(n\geq m\)
• B. \({}^{n}P_{m}\), if \(n\geq m\)
• C. \(0\), if \(n\geq m\)
• D. \(m\cdot {}^{n}C_{m}\), if \(n\geq m\)

Hint:

For injections from a set with \(m\) elements to a set with \(n\) elements: \[ \text{Number of injections}={}^{n}P_{m} \] because distinct images are required for every element.

Answer 1

The Correct Option is B

Step 1: Recall what an injection means.
An injection (one-one map) from $A$ to $B$ sends distinct elements of $A$ to distinct elements of $B$, so no repetition of images is allowed.
Step 2: Note the sizes.
Here $|A|=m$ and $|B|=n$ with $n\ge m$ so that enough distinct images exist.
Step 3: Count choices one element at a time.
The first element of $A$ has $n$ choices, the second has $n-1$, the third $n-2$, and so on.
Step 4: Multiply the choices.
Total $=n(n-1)(n-2)\cdots(n-m+1)$, a product of $m$ falling factors.
Step 5: Recognise the permutation form.
This product equals $\dfrac{n!}{(n-m)!}={}^{n}P_{m}$.
Step 6: Conclude.
So the number of injections is ${}^{n}P_{m}$ for $n\ge m$, which is option (2).
\[ \boxed{{}^{n}P_{m},\ \text{if } n\ge m} \]