Explanation

(a)
For the first player, distribute the cards in $\ ^{52}C_{17}$ ways. Now, out of 35 cards left 17 cards can be put for second player in $\ ^{35}C_{17}\$ways. Similarly, for third player put them in $\ ^{18}C_{17}$ways. One card for the last player can be put in $\ ^{1}C_{1}$ way. Therefore, the required number of ways for the proper distribution
$= \ ^{52}C_{17} \times \ ^{35}C_{17} \times \ ^{18}C_{17} \times \ ^{1}C_{1}$
$= \frac{52!}{35!17!} \times \frac{35!}{18!17!} \times \frac{18!}{17!1!} \times 1! = \frac{52!}{(17!)^{3}}$