Question

2. How many seven – letter code words can be formed using a standard 26 letter alphabet (12 points)

a.         if repetition is allowed?

b.         if repetition is not allowed?

c.         if the first letter has to be a C and the rest of the letters are different?

d.         if the first two letters have to be a vowel and the rest of the letters are different?

Answer 1

Answer This is a permutation Problem

How many seven letter code words can be formed using 26 letter alphabet

a> if repetition is allowed.

in this case 1st letter can be filled by 26 ways , 2nd letter can be selected in 25 ways so on.

So no of options =26*25*24*23*22*21*20 = 26! */19! = 26P7 = PERMUT(26,7)= 3315312000

Note !-Factorial sign e.g 3! = 3*2*1 =6(permut function in xl )

b) if repetition is allowed . Each place can be filled 26 ways . Hence total no of code word = 26⁷=8031810176

c)  if the first letter has to be a C and the rest of the letters are different?

in this case 1st letter can be done in 1 way, for rest 6 letters need be decidedf from 25 (albhate-c)  

Hence no of codes will be 25P6 = 127512000

d) if the first two letters have to be a vowel and the rest of the letters are different?

1st two letters have to be vowel (assuming repetion not allowed ) can be selected in 5P2= 5*4 = 20 ways

Rest 5 letters need to be selected from 21 letter 21P5 = 2441880

Hence total no of options 21P5 * 5P2 = 2441880 *20 =48837600 ways