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

<p>$\textbf{Step 1: Total Functions with } 1 \in f(A)$</p> <p>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</p> <p>$4^4 = 256.$</p> <p>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</p> <p>$3^4 = 81.$</p> <p>Thus, the number of functions such that $ 1 \in f(A) $ is</p> <p>$256 - 81 = 175.$</p> <p>$\textbf{Step 2: Counting Many-One Functions}$</p> <p>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</p> <p>$4! = 24.$</p> <p>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) $.</p> <p>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 $:</p> <p>$175 - 24 = 151.$</p> <p>$\boxed{151}$</p> <p>This detailed explanation shows that the number of many-one functions $ f: A \rightarrow B $ such that $ 1 \in f(A) $ is indeed $ 151 $.</p>