Question

The total number of injective mappings from a set with m elements to a set with n elements, $\mathrm{m}\le \mathrm{n}$, is

• A. ${\mathrm{m}}^{\mathrm{n}}$
• B. ${\mathrm{n}}^{\mathrm{m}}$
• C. $\frac{\mathrm{n}!}{(\mathrm{n}-\mathrm{m})!}$
• D. $\text{n!}$

Answer 1

The correct option is C

$\frac{\mathrm{n}!}{(\mathrm{n}-\mathrm{m})!}$

Explanation for the correct option:

Option (C):

Given data:

• First set with n elements and another set with m elements, where, $n⩾m$.
• Now we have to find out the total number of injective mappings from a set with m elements to a set with n elements.

Injective mapping

• As we all know that injective mapping is also known as one – to – one function i.e. a function that maps distinct elements of its domain to distinct elements of its co-domain, or we can say that every element of its co-domain is the image of at most one element of its domain.
• Now we have to map from a set with m elements to a set with n elements.
  So first find out the number of ways to select m elements from n elements as, $n⩾m$.
• Now as we know that if there are n objects so the number of ways to select r objects out of n is ${}^n{C}_r$.
  So the number of ways to select m elements from n elements is ${}^n{C}_m$.
• Now we have to arrange these elements so the number of ways to arrange m elements is m!.
  So the total number of injective mapping is the product of the above two values.
• So the total number of injective mapping =${}^n{C}_m\left(m!\right)$
  Now as we know that,${}^n{C}_r={\displaystyle \frac{n!}{r!\left(n-r\right)!}}$
  Therefore, the total number of injective mapping =${\displaystyle \frac{n!}{m!\left(n-m\right)!}}\left(m!\right)={\displaystyle \frac{n!}{\left(n-m\right)!}}$
  So this is the required answer.
  Hence option (c) is the correct answer

Hence option (c) is the correct answer.