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).

Solutions

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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