If n(A)=4, n(B)=5 and number of functions from A to B such that range contains exactly 3 elements is k, `k/(60)` is ………. .

Answer

To solve the problem, we need to find the number of functions from set A to set B such that the range contains exactly 3 elements. Here are the steps to arrive at the solution:

### Step 1: Identify the sizes of sets A and B
Given:
- \( n(A) = 4 \) (the number of elements in set A)
- \( n(B) = 5 \) (the number of elements in set B)

### Step 2: Choose 3 elements from set B
We need to select exactly 3 elements from set B to be in the range of the function. The number of ways to choose 3 elements from a set of 5 is given by the combination formula:

\[ \text{Number of ways to choose 3 elements from 5} = \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10 \]

### Step 3: Map elements from A to the selected 3 elements in B
Now, we need to map the 4 elements from set A to the 3 selected elements in set B such that at least one of the selected elements in B is mapped to by exactly 2 elements from A.

#### Step 3.1: Choose 2 elements from A to map to one of the 3 selected elements
We can choose 2 elements from the 4 elements in A to map to one of the selected 3 elements in B. The number of ways to choose 2 elements from 4 is:

\[ \text{Number of ways to choose 2 elements from 4} = \binom{4}{2} = \frac{4!}{2!(4-2)!} = \frac{4 \times 3}{2 \times 1} = 6 \]

#### Step 3.2: Choose which of the 3 selected elements will receive the 2 mappings
For each pair of elements chosen from A, we can choose any of the 3 selected elements in B to map them to. This gives us 3 choices.

#### Step 3.3: Map the remaining 2 elements from A to the remaining 2 elements in B
The remaining 2 elements from A can be mapped to the remaining 2 elements in B in \( 2! \) (factorial of 2) ways:

\[ 2! = 2 \]

### Step 4: Calculate the total number of functions
Now, we can combine all these choices together:

\[ \text{Total number of functions} = \text{(ways to choose 3 elements from B)} \times \text{(ways to choose 2 elements from A)} \times \text{(ways to choose which B element gets 2 mappings)} \times \text{(ways to map remaining A elements)} \]

Putting it all together:

\[ k = \binom{5}{3} \times \binom{4}{2} \times 3 \times 2 = 10 \times 6 \times 3 \times 2 \]

Calculating this:

\[ k = 10 \times 6 = 60 \]
\[ 60 \times 3 = 180 \]
\[ 180 \times 2 = 360 \]

### Step 5: Find \( \frac{k}{60} \)
Now, we need to find \( \frac{k}{60} \):

\[ \frac{k}{60} = \frac{360}{60} = 6 \]

Thus, the final answer is:

\[ \boxed{6} \]