1.1.10 problem 12

(a)

$$\left(\frac{52}{13}\right)$$

(b)

The number of ways to break $52$ cards into $4$ groups of size $13$ is $$\frac{\left(\frac{52}{13}\right)\left(\frac{39}{13}\right)\left(\frac{26}{13}\right)\left(\frac{13}{13}\right)}{4!}$$

.

The reason for dividision by $4!$ is that all permutations of specific $4$ groups describe the same way to group $52$ cards.

Since we do care about the order of the $4$ groups, we should not divide by $4!$. The final answer then is

$$\left(\frac{52}{13}\right)\left(\frac{39}{13}\right)\left(\frac{26}{13}\right)\left(\frac{13}{13}\right)$$

(c)

The key is to notice that the sampling is done without replacement. ${\left(\frac{52}{13}\right)}^4$ assumes that all four players have $\left(\frac{52}{13}\right)$ choices of hands available to them. This would be true if sampling was done with replacement.