(d) \(\frac{52!}{(16!)^3\times(3!)}\)
First player can get 16 cards in 52C16 ways
Second player can get 16 cards in 36C16 ways
Third player can get 16 cards in 20C16 ways
Fourth player can get 4 cards in 4C4 ways
But the first three sets can be interchanged in 3! ways
∴ Required number of ways = 52C16 × 36C16 × 20C16 × 4C4 × \(\frac1{3!}\)
= \(\frac{52!}{(16!)^3\times(3!)}\)