Let A=1,2,3,4,5 and B=1,2,3,4,5,6.

Question

Let $A={1,2,3,4,5}$ and $B={1,2,3,4,5,6}$. Then the number of functions $f:A	o B$ satisfying $f(1)+f(2)=f(4)-1$ is equal to \_\_\_\_\_\_\_\_\_\_.

Answer: 360

Accepted Answer

Given that the function $f:A	o B$ satisfies the condition $f(1)+f(2)=f(4)-1$, where the set $A={1,2,3,4,5}$ and the set $B={1,2,3,4,5,6}$.

We want to find out how many such functions exist.

First, observe that the condition $f(1)+f(2)=f(4)-1$ can be rewritten as $f(1)+f(2)+1=f(4)$. So, the sum of $f(1),f(2),$ and 1 is equal to $f(4)$. Since $f(4)$ is a value in set B, it can take values from 1 to 6.

The maximum value of $f(1)+f(2)+1$ can be $6+6+1=13$, but this is more than 6 (the maximum value of $f(4)$), so it's not possible. Thus, the maximum value of $f(4)$ in this case can be 6.

Let's now analyze the number of functions for each value of $f(4)$ from 3 to 6 (we start from 3 because $f(1)$ and $f(2)$ take values from set B and their minimum sum plus 1 is 3):

1. When $f(4)=6$, then $f(1)+f(2)=5$. The pairs $(f(1),f(2))$ that satisfy this equation are $(1,4),(2,3),(3,2),(4,1)$. For each of these 4 cases, the functions $f(3)$ and $f(5)$ can each take 6 values from set B, resulting in $4	imes 6	imes 6=144$ functions.

2. When $f(4)=5$, then $f(1)+f(2)=4$. The pairs $(f(1),f(2))$ that satisfy this equation are $(1,3),(2,2),(3,1)$. For each of these 3 cases, the functions $f(3)$ and $f(5)$ can each take 6 values from set B, resulting in $3	imes 6	imes 6=108$ functions.

3. When $f(4)=4$, then $f(1)+f(2)=3$. The pairs $(f(1),f(2))$ that satisfy this equation are $(1,2),(2,1)$. For each of these 2 cases, the functions $f(3)$ and $f(5)$ can each take 6 values from set B, resulting in $2	imes 6	imes 6=72$ functions.

4. When $f(4)=3$, then $f(1)+f(2)=2$. The only pair $(f(1),f(2))$ that satisfies this equation is $(1,1)$. For this case, the functions $f(3)$ and $f(5)$ can each take 6 values from set B, resulting in $1	imes 6	imes 6=36$ functions.

Adding the numbers of functions from all these cases, we get a total of $144+108+72+36=360$ functions from $A$ to $B$ that satisfy the given condition.

Therefore, the number of functions $f:A	o B$ satisfying the condition $f(1)+f(2)=f(4)-1$ is 360.

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