Let R = { a, b, c, d, e } and S = {1, 2, 3, 4 | vedclass

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(A) The total number of onto functions from a set with $n=5$ elements to a set with $m=4$ elements is given by the formula $m! 	imes S_2(n, m)$,where

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$180$

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(A) The total number of onto functions from a set with $n=5$ elements to a set with $m=4$ elements is given by the formula $m! 	imes S_2(n, m)$,where $S_2(n, m)$ is the Stirling number of the second kind.Total onto functions = $4! 	imes S_2(5, 4) = 24 	imes \binom{5}{2} = 24 	imes 10 = 240$.Now,we calculate the number of onto functions where $f(a) = 1$. If $f(a) = 1$,then the remaining $4$ elements ${b, c, d, e}$ must map onto ${1, 2, 3, 4}$ such that the function remains onto. This means the set ${b, c, d, e}$ must map onto ${2, 3, 4}$ (surjective) or map onto ${1, 2, 3, 4}$ (surjective).Case $1$: $f(a)=1$ and the range is ${1, 2, 3, 4}$. The remaining $4$ elements must map to ${1, 2, 3, 4}$ such that one element maps to $1$ and the others map to ${2, 3, 4}$ in a way that covers all values. This is equivalent to the number of onto functions from a $4$-element set to a $4$-element set,which is $4! = 24$.Case $2$: $f(a)=1$ and the range is ${2, 3, 4}$. The remaining $4$ elements must map onto ${2, 3, 4}$. The number of onto functions from a $4$-element set to a $3$-element set is $3! 	imes S_2(4, 3) = 6 	imes \binom{4}{2} = 6 	imes 6 = 36$.Total functions where $f(a) = 1$ is $24 + 36 = 60$.Therefore,the number of onto functions where $f(a) 
eq 1$ is $240 - 60 = 180$.

Let $R = \{ a, b, c, d, e \}$ and $S = \{1, 2, 3, 4\}$. The total number of onto functions $f: R \rightarrow S$ such that $f(a) \neq 1$ is equal to $.............$.

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