Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • How many different four-letter codes can be formed using the letters A, B, C, D, and E if repetition is allowed and the codes should have starting letter D?  <div class='qna-opti

How many different four-letter codes can be formed using the letters A, B, C, D, and E if repetition is allowed and the codes should have starting letter D?

 

Option: 1

125


Option: 2

150


Option: 3

100


Option: 4

200


Answers (1)

best_answer

If repetition is allowed and the codes should have the starting letter D, we can treat the starting letter as fixed and consider the remaining three positions.

For the second, third, and fourth positions, we have 5 choices (A, B, C, D, or E) since repetition is allowed.

Therefore, the total number of different four-letter codes with repetition allowed and starting letter D is calculated as:

1 (choice for the starting letter D) \times5 (choices for the second position) \times5 (choices for the third position)\times 5 (choices for the fourth position) = 1\times 5\times 5 \times5 = 125.

Hence, there are 125 different four-letter codes that can be formed using the letters A, B, C, D, and E, where repetition is allowed and the codes have the starting letter D.

 

Posted by

Kuldeep Maurya

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main Maths 2025: Five tips to score high marks in ‘toughest’ subject; most important chapters
anam-khan

JEE Main 2025 exam city intimation slip for session 1 soon at jeemain.nta.nic.in
anam-khan

Focus on modern physics, formulas to solve tricky problems; 5 tips to tackle JEE Main physics 2025

Latest Question