Question

A person has to create a code. The code must contain exactly 7 letters. The first letter must either a N, P, R, or T. The second letter must be a vowel that is not I, O, or U. The third letter must be a consonant that is not a D, F, H, or J. The fourth letter must be a vowel that is not an O or U. The fifth letter must be a consonant that is not a D or F. The sixth 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

Answer 1

ANSWER::

Since Repetition is not allowed, a letter once fixed in a place cannot come in any other position

There are a Total of 26 alphabets of which 5 are vowels (A, E, I , O and U), and the remaining 21 are consonants.

___________________________________________________

The first letter can be any of the 4 of N, P, R or T (All 4 are consonants).

(So one consonant is removed from the 21.)

Therefore number of ways of choosing a letter for the 1st place = 4

____

The second letter must not be an I, O or U, which means the second place can be A or E. (One Vowel gets removed here).

Therefore number of ways of choosing a letter for the 2nd place = 2

___

The third letter should not be D, F, H or J. All are consonants. We now have 21 - 1 - 4 consonants to choose from for the third place as one of N, P, R or T has been fixed in the first place. (In this place one more consonant gets removed).

Therefore number of ways of choosing a letter for the 3rd place = 16

___

The fourth letter must be a vowel that is not an O or U, that is it must be an A, E or I.

Here 2 different cases arises.

Case (i) The 2nd letter was an A, then the number of ways of choosing the 4th place = 2

Case (ii) The 2nd was not an A, then the number of ways of choosing the 4th place = 3

____

The Fifth Letter should not have D, F, H or J. All are consonants. We now have 21 - 1 - 1 - 2 consonants to choose from for the third place as one of N, P, R or T has been fixed in the first place and one is fixed in the 3rd place . (One more consonant gets removed in this place).

Therefore number of ways of choosing a letter for the 3rd place = 20

___

The Sixth letter has to be a consonant. We now have 21 - 1 - 1 - 1 = 18 consonants available for the sixth place (as 3 consonants have gone into the 1st, 3rd and 5th place)

______

The seventh letter just has to be an alphabet. There are 26 - 6 = 20 alphabets free for the last place

__________________________________

Since the 4th letter is composed of 2 different events, we have

(i) Letters having A in the 4th place = 4 * 2 * 16 * 2 * 20 * 18 * 20 = 1843200

(ii) Letters not having A in the 4th place = 4 * 2 * 16 * 3 * 20 * 18 * 20 = 2764800

Therefor the total number of ways = 1843200 + 2764800 = 4,608,000 codes

(OR) TRY THIS ANSWER

NOTE:: I HOPE THIS ANSWER IS HELPFULL TO YOU......**PLEASE SUPPORT ME WITH YOUR RATING......

**PLEASE GIVE ME "LIKE".....ITS VERY IMPORTANT  FOR,ME......PLEASE SUPPORT ME .......THANK YOU