If $$\binom{30}{1}^2 + 2\binom{30}{2}^2 + 3\binom{30}{3}^2 + \ldots + 30\binom{30}{30}^2 = \frac{\alpha \cdot 60!}{(30!)^2}$$, then $$\alpha$$ is equal to

$$S = 1 \cdot \binom{30}{1}^2 + 2 \cdot \binom{30}{2}^2 + 3 \cdot \binom{30}{3}^2 + \dots + 30 \cdot \binom{30}{30}^2$$

$$S = \sum_{r=1}^{30} r \cdot \binom{30}{r}^2$$

Using the property $$\binom{30}{r} = \binom{30}{30-r}$$:

$$S = \sum_{r=1}^{30} \left[ r \cdot \binom{30}{r} \cdot \binom{30}{30-r} \right]$$

Now substitute $$r \cdot \binom{30}{r} = 30 \cdot \binom{29}{r-1}$$:

$$S = \sum_{r=1}^{30} \left[ 30 \cdot \binom{29}{r-1} \cdot \binom{30}{30-r} \right]$$

$$S = 30 \sum_{r=1}^{30} \binom{29}{r-1} \cdot \binom{30}{30-r}$$

$$\sum_{r=1}^{30} \binom{29}{r-1} \cdot \binom{30}{30-r} = \binom{29+30}{29} = \binom{59}{29}$$

$$S = 30 \cdot \binom{59}{29}$$

$$S = 30 \cdot \frac{59!}{29! \cdot (59-29)!} = 30 \cdot \frac{59!}{29! \cdot 30!}$$

$$S = \frac{1}{2} \cdot \frac{60!}{29! \cdot 30!}$$

$$S = \frac{30}{2} \cdot \frac{60!}{30! \cdot 30!} = 15 \cdot \frac{60!}{(30!)^2}$$

$$\alpha = 15$$