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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: littledreamer on January 18, 2016, 08:07:54 pm
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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
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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 :D