0 0 votes Let $A$ and $B$ be finite sets with $|A|=7$ and $|B|=4$. Find the number of onto functions $f:A\to B$ such that exactly two elements of $B$ have exactly one preimage each. Set Theory & Algebra discrete-mathematics goclasses goclasses-cs-dpp goclasses-cs-dpp-day-265 goclasses-dm-practice-questions functions numerical-answers + – GO Classes 108 views answer comment Share Follow Print 0 reply Please log in or register to add a comment.
0 0 votes First, we handle the condition that exactly two elements in set $B$ have exactly one preimage in set $A$.Choose the targets in $B:$ There are $4$ elements in $B$. We need to choose $2$ of them to have exactly one preimage.$$\binom{4}{2} = \frac{4!}{2!(4-2)!} = 6 \text{ ways}$$Choose the preimages in $A:$ There are $7$ elements in $A$. We need to choose $2$ of them to map to the $2$ selected elements in $B$.$$\binom{7}{2} = \frac{7!}{2!(7-2)!} = 21 \text{ ways}$$Map them : We have $2$ chosen elements in $A$ and $2$ chosen elements in $B$. There are $2! = 2$ ways to uniquely map them to each other.For this first phase, there are $6 \times 21 \times 2 = 252$ possible combinations. Now, we have $7 - 2 = 5$ elements remaining in $A$ and $4 - 2 = 2$ elements remaining in $B$.For the function to be onto, these remaining $5$ elements in $A$ must map to the remaining $2$ elements in $B$. Furthermore, to satisfy the rule that exactly two elements in $B$ have one preimage, neither of these two remaining elements in $B$ can end up with exactly one preimage.Let's find the number of valid mappings:The total number of onto functions from a set of $5$ elements to a set of $2$ elements is $2^5 - 2 = 30$.We must exclude any mappings where one of the targets gets exactly $1$ preimage (which means the other gets $4$).The number of ways to choose $1$ element from the remaining $5$ to be the sole preimage is $\binom{5}{1} = 5$. Since there are $2$ target elements in $B$ it could map to, there are $5 \times 2 = 10$ invalid onto functions.Subtract the invalid functions from the total onto functions: $30 - 10 = 20$ valid ways.Alternatively, we can think of this as partitioning the $5$ remaining elements of $A$ into two sets of size $3$ and $2$.Ways to partition: $\binom{5}{3} = 10$.Ways to assign these partitions to the $2$ remaining elements in $B: 2! = 2$.$10 \times 2 = 20$ valid ways. Therefore, the total number of onto functions is $ = 252 \times 20 = \boxed{\mathbf{5040}}$ GO Classes answered May 7 GO Classes comment Share Follow 0 reply Please log in or register to add a comment.