A person has to create a code. The code must contain exactly 6 letters. The first...

A person has to create a code. The code must contain exactly 6 letters. The first letter must be a vowel that is not an E or I. The second letter must be either a J, K, or L. The third letter must be a consonant that is not a G or R. The fourth letter must be a vowel that is not an I. The fifth letter must be a consonant. If repetition of letters are not allowed, how many different codes can the person make? (Note: a vowel is A, E, I, O, or U, and consonants are letters that are not vowels).

Expert Solution

There are five vowels A,E,I,O,U and 21 consonants,

The first letter has 3 options (A,O,U) to choose since other vowels E or I cant be chosen.

The second letter has 3 options to chosen from J,K,or L .

The third letter has 21 - 2 =19 options to choose since G and R cant be used so we subtracted 2 consonants from 21.

Now there will be one consonant (either J,K or L) which will be used in second letter formation , since repetition is not allowed , we have finally 19 - 1 = 18 options to choose for third letter .

The fourth letter should be vowel , but w have used one vowel for first letter from A,O,U. so we have 5 -1 =4 options to choose, there is condition given that I cant be used , so finally we will have 4 -1 = 3 options from to choose for the fourth letter.

For second letter and fourth letter we have used 2 consonants, so we have 21 - 2 =19 consonants to choose for the fifth letter.

There is no mention of any conditions for 6th letter , since 5 letters have already been used we have 26 - 5 = 21 options for the 6th letter.

Number of different codes that can be formed = 3 X 3 X 18 X 3 X 19 X 21 = 193914

orchestra answered 1 year ago

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