Number of One to One and Onto Functions Between Finite Sets

To find the required value of $\frac{1}{5!} (\beta - \alpha)$, we evaluate the values of $\alpha$ and $\beta$ individually.

Step 1: Calculate $\alpha$

$\alpha$ is defined as the number of one-one (injective) functions from set $X$ to set $Y$, where $|X| = 5$ and $|Y| = 7$.

The number of one-one functions from a set with $5$ elements to a set with $7$ elements is given by the permutation ${^7P_5}$: $\alpha = {^7P_5} = \frac{7!}{(7-5)!} = 7 \times 6 \times 5 \times 4 \times 3 = 2520$

Dividing $\alpha$ by $5!$, we get: $\frac{\alpha}{5!} = \frac{^7P_5}{5!} = {^7C_5} = \frac{7 \times 6}{2 \times 1} = 21$

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Step 2: Calculate $\beta$

$\beta$ is defined as the number of onto (surjective) functions mapping set $Y$ to set $X$, where $|Y| = 7$ and $|X| = 5$.

The number of onto functions from a set of size $n$ to a set of size $k$ is given by $k! \cdot S(n, k)$, where $S(n, k)$ is the Stirling number of the second kind, representing the number of ways to partition $n$ distinct elements into $k$ non-empty sets.

Here, $n = 7$ and $k = 5$. The partitions of $7$ elements into $5$ non-empty groups can occur in two possible patterns:

1. Partition of type $(3, 1, 1, 1, 1)$: $\text{Number of ways} = \binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$

2. Partition of type $(2, 2, 1, 1, 1)$: $\text{Number of ways} = \frac{1}{2!} \binom{7}{2} \binom{5}{2} = \frac{1}{2} \times 21 \times 10 = 105$

Summing both partition types gives the total Stirling number of the second kind $S(7, 5)$: $S(7, 5) = 35 + 105 = 140$

Therefore, the total number of onto functions $\beta$ is: $\beta = 5! \cdot S(7, 5) = 5! \times 140$

Dividing $\beta$ by $5!$, we get: $\frac{\beta}{5!} = 140$

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Step 3: Compute $\frac{1}{5!} (\beta - \alpha)$

Substitute the values derived above: $\frac{1}{5!} (\beta - \alpha) = \frac{\beta}{5!} - \frac{\alpha}{5!} = 140 - 21 = 119$

The final value is 119.