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A code is to be made by arranging 7 letters. Three of the letters used will be the letter A, two of the letters used will be the letter B, one of the letters used will be the letter C, and one of the letters used will be the letter D. If there is only one way to present each letter, how many different codes are possible?
  • a)
    42
  • b)
    210
  • c)
    420
  • d)
    840
  • e)
    5040
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
A code is to be made by arranging 7 letters. Three of the letters used...
Step 1: Analyze the Question :- We have to make a seven-letter code, but some of our letters are repeated. We have three As two Bs, one C, and one D. We have to calculate the possible number of different codes.
Step 2: State the Task :- We'll calculate the number of permutations, remembering to take the repeated letters into account.
Step 3: Approach Strategically :- To calculate the number of permutations where some of the elements are indistinguishable, we'll divide the total number of permutations by the factorial of the number of indistinguishable elements.
So we have :- 
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Most Upvoted Answer
A code is to be made by arranging 7 letters. Three of the letters used...
7factorial/ 3factorial into 2factorial and here 7 7 factorial is total no of possible codes without repeating letter then we will devide it by the factorial of no of repeated letters
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Community Answer
A code is to be made by arranging 7 letters. Three of the letters used...
Explanation:

Arranging the letters:
- There are 7 letters in total to be arranged.
- Out of these 7 letters, 3 are A's, 2 are B's, 1 is C, and 1 is D.
- Therefore, the total number of ways to arrange these 7 letters is 7!/(3! * 2! * 1! * 1!) = 420.
Therefore, the correct answer is option C) 420.
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