Let `X` be a set with exactly 5 elements and `Y` be a set with exactly 7 elements. If `alpha` is the number of one-one function from `X` to `Y` and `beta` is the number of onto function from `Y` to `X` , then the value of `1/(5!)(beta-alpha)` is _____.

Explanation

To solve the problem, we need to find the values of \( \alpha \) (the number of one-one functions from set \( X \) to set \( Y \)) and \( \beta \) (the number of onto functions from set \( Y \) to set \( X \)), and then compute the expression \( \frac{1}{5!}(\beta - \alpha) \).

### Step 1: Calculate \( \alpha \) (Number of one-one functions from \( X \) to \( Y \))

1. **Understanding one-one functions**: A one-one function means that each element of \( X \) must map to a unique element in \( Y \). Since \( X \) has 5 elements and \( Y \) has 7 elements, we can choose any 5 elements from \( Y \) to map to the elements of \( X \).

2. **Choosing elements from \( Y \)**: The number of ways to choose 5 elements from 7 is given by the combination formula:
\[ \binom{7}{5} = \frac{7!}{5! \cdot (7-5)!} = \frac{7!}{5! \cdot 2!} = \frac{7 \times 6}{2 \times 1} = 21 \]

3. **Arranging the chosen elements**: Once we have chosen 5 elements from \( Y \), we can arrange them in \( 5! \) ways. Therefore, the total number of one-one functions \( \alpha \) is:
\[ \alpha = \binom{7}{5} \times 5! = 21 \times 120 = 2520 \]

### Step 2: Calculate \( \beta \) (Number of onto functions from \( Y \) to \( X \))

1. **Understanding onto functions**: An onto function means every element of \( X \) must be mapped to by at least one element of \( Y \).

2. **Total functions from \( Y \) to \( X \)**: The total number of functions from \( Y \) (7 elements) to \( X \) (5 elements) is:
\[ 5^7 \]

3. **Using the principle of inclusion-exclusion**: We need to subtract the cases where at least one element of \( X \) is not mapped to. We can calculate the number of onto functions using the formula:
\[ \beta = \sum_{k=0}^{5} (-1)^k \binom{5}{k} (5-k)^7 \]
where \( k \) is the number of elements in \( X \) that are not used.

- For \( k = 0 \): \( \binom{5}{0} \cdot 5^7 = 1 \cdot 78125 = 78125 \)
- For \( k = 1 \): \( \binom{5}{1} \cdot 4^7 = 5 \cdot 16384 = 81920 \)
- For \( k = 2 \): \( \binom{5}{2} \cdot 3^7 = 10 \cdot 2187 = 21870 \)
- For \( k = 3 \): \( \binom{5}{3} \cdot 2^7 = 10 \cdot 128 = 1280 \)
- For \( k = 4 \): \( \binom{5}{4} \cdot 1^7 = 5 \cdot 1 = 5 \)
- For \( k = 5 \): \( \binom{5}{5} \cdot 0^7 = 1 \cdot 0 = 0 \)

Now, applying inclusion-exclusion:
\[ \beta = 78125 - 81920 + 21870 - 1280 + 5 - 0 = 16800 \]

### Step 3: Calculate \( \frac{1}{5!}(\beta - \alpha) \)

1. **Substituting the values**:
\[ \beta - \alpha = 16800 - 2520 = 14280 \]

2. **Calculating \( \frac{1}{5!}(\beta - \alpha) \)**:
\[ \frac{1}{5!}(14280) = \frac{14280}{120} = 119 \]

### Final Answer
The value of \( \frac{1}{5!}(\beta - \alpha) \) is \( \boxed{119} \).