SOLUTION: 2. If the permutation of the word WHITE is selected at random, how many of the permutations i. Begins with a consonant? ii. Ends with a vowel? iii. Has a consonant and vowels al

Question 1195966: 2. If the permutation of the word WHITE is selected at random, how many of the permutations
i. Begins with a consonant?
ii. Ends with a vowel?
iii. Has a consonant and vowels alternating?

Answer by ikleyn(52754) (Show Source): You can put this solution on YOUR website!
.
If the permutation of the word WHITE is selected at random, how many of the permutations
i. Begins with a consonant?
ii. Ends with a vowel?
iii. Has a consonant and vowels alternating?
~~~~~~~~~~~~~~~~~~~

(i)   The total number of permutations of the letters W,H,I,T,E is 5! = 5*4*3*2*1 = 120.

      Of them, the number of permutations starting with the consonant W is 4! = 4*3*2*1 = 24;
          
               the number of permutations starting with the consonant H is another 4! = 4*3*2*1 = 24;

               the number of permutations starting with the consonant T is another 4! = 4*3*2*1 = 24.

       The total number of permutations starting with the consonant is 3*24 = 72.    ANSWER



(ii)   Similar reasoning with the last letters gives the number of permutations for question (ii)  2*24 = 48.    ANSWER



(iii)  For alternating locations, we have these permutations  CVCVC, where C is a place holder  

       for any of the three consonants and V is a place holder for any of the two participating vowels.


       So, the number of the alternate permutations is 3!*2! = 6*2 = 12.       ANSWER

Solved.

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