Problem 3

Problem – 3

How many strings of six uppercase English letters are there

a) if letters can be repeated?

b) if no letter can be repeated?

c) that start with Y, if letters can be repeated?

d) that start with Y, if no letter can be repeated?

e) that start and end with Y, if letters can be repeated?

f ) that start with the letters ER (in that order), if letters can be repeated?

g) that start and end with the letters ER (in that order), if letters can be repeated?

h) that start or end with the letters ER (in that order), if letters can be repeated?

 

 

Solution:

a)      There are possible 26 letters to fill the six positions, repetition is allowed so we can fill every place with 26 letters,  possible strings are = 26x26x26x26x26x26 = 308915776

b)      There are possible 26 letters to fill the six positions, repetition is not allowed so we can fill first place with 26 letters and second with 25 letters and so on,  possible strings are = 26x25x24x23x22x21 = 165765600

c)      We have to fill first place with Y letter, and rest 5 places with 26 letters as repetition is allowed, so number of possible strings are = 26x26x26x26x26 = 11881376

d)      We have to fill first place with Y letter, and 2nd with 25 letters and 3rd with 24 letters and so on, as repetition is not allowed, so number of possible strings are = 25x24x23x22x21 = 6375600

e)      We have to fill first and last place with Y letter and rest 4 with 26 letters as repetition is allowed. So number of possible strings are = 26x26x26x26 = 456976

f)       First two places are filled by letters ER and rest 4 are filled by 26 letters for each place as repetition is allowed, so number of possible strings are =26x26x26x26 = 456976

g)      First two and last two places are filled by letters ER, and rest 2 places are filled by 26 letters for each place, so number of possible strings are =26×26 = 676

h)      When string start with ER than possible number of strings are 456976 ( from part e ) , and when string ends with ER than possible number of strings are 456976. So total number of strings will be 456976+456976 = 913952

It included the strings which have ER at starting and ending positions , so there 676 positions which have ER are at starting and ending positions ( from part g) so we have to minus that positions to get the actual possible strings = 913952 – 676 = 913276

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