Counting methods question, please help

[littledreamer]

From a 1/2 Methods textbook:
There are 5 vowels and 21 consonants in the English alphabet. How many different four-letter ‘words’ can be formed that contain two different vowels and two different consonants?

Answer is 50400

Thanks in advance

[Peanut Butter]

From a 1/2 Methods textbook:
There are 5 vowels and 21 consonants in the English alphabet. How many different four-letter ‘words’ can be formed that contain two different vowels and two different consonants?

Answer is 50400

Thanks in advance

Use the box method

5 (vowels) x 4 (vowels excluding vowel already used) x 21 (consonants) x 20 (consonants excluding consonant already used) = 8400

Therefore, 8400 possible arrangements for words in this order

We then multiply 8400 by the number of different possible orders.

Possible orders: VVCC, VCVC, VCCV, CCVV, CVCV, CVVC

There are six possible orders.

Therefore, 8400 x 6 = 50 400

Hope that makes sense