Let A=1,2,3,4 and B=1,4,9,16.

Question

Let $A={1,2,3,4}$ and $B={1,4,9,16}$. Then the number of many-one functions $f:A	o B$ such that $1\in f(A)$ is equal to :

Accepted Answer

$$ \textbf{ Total Functions with } 1 \in f(A) $$

Any function $ f: A \to B $ where $ A = \{1, 2, 3, 4\} $ and $ B = \{1, 4, 9, 16\} $ is defined by choosing one of the four elements of $ B $ for each element of $ A $. Thus, the total number of functions is

$4^4 = 256.$

To count those functions where $ 1 $ appears at least once in the set $ f(A) $, we can use the complementary counting method: subtract the functions that never use $ 1 $. If $ 1 $ is excluded, each element of $ A $ has only 3 choices (namely, $ \{4, 9, 16\} $), so the number of such functions is

$3^4 = 81.$

Thus, the number of functions such that $ 1 \in f(A) $ is

$256 - 81 = 175.$

$$ \textbf{ Counting Many-One Functions} $$

In this context, "many-one functions" are understood to be non-injective functions. Since an injective (one-to-one) function from $ A $ to $ B $ must be a permutation (because both sets have 4 elements), the number of one-to-one functions is

$4! = 24.$

It is important to note that every injective function $ f: A \to B $ has $ f(A) = B $ (a full permutation) which automatically means $ 1 \in f(A) $.

Thus, the number of many-one (non-injective) functions $ f: A \to B $ with $ 1 \in f(A) $ is found by subtracting the one-to-one functions from the total functions that include $ 1 $:

$175 - 24 = 151.$

$$ \boxed{151} $$

This detailed explanation shows that the number of many-one functions $ f: A \rightarrow B $ such that $ 1 \in f(A) $ is indeed $ 151 $.

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