The English alphabet has 5 vowels and 21 consonants.

2 different vowels and 2 different consonants are to be selected from the English alphabet.

Since there are 5 vowels in the English alphabet, number of ways of selecting 2 different vowels from the alphabet $={ }^{5} \mathrm{C}_{2}=\frac{5 !}{2 ! 3 !}=10$

Since there are 21 consonants in the English alphabet, number of ways of selecting 2 different consonants from the alphabet $={ }^{21} C_{2}=\frac{21 !}{2 ! 19 !}=210$

Therefore, number of combinations of 2 different vowels and 2 different consonants = 10 × 210 = 2100

Each of these 2100 combinations has 4 letters, which can be arranged among themselves in 4! ways.

Therefore, required number of words = 2100 × 4! = 50400