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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?
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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...
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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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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This is a typical permutation-probability problem. To make this problem easily understandable we will break into two parts: i. First, we will find out all the seven lettered words from the letters of word CLASSIC ii. Next, we will find out how many of these words will have the two Cs together. The total number of words formed using the seven letters from the word CLASSIC is found by using the multiplication principle. There are seven places for each of the seven letters. The first place has 7 choices, the second place has (7-1) =6 choices, the third places has 5 choices and the seventh place has 1 choice. Hence, the total number of words formed is: = 7 x 6 x 5 x 4 x ... x 1 = 7! Notice that there are two Cs and two S in the word, which can be treated as repeated elements. To adjust for the repeated elements we will divide 7! by the product of 2! x 2! So, the total number of words formed is: 7!/(2! x 2!) We need to find how many of these words will have the two Cs together. To do this, let us treat the two Cs as a single entity. So, now we have six spaces to fill. Continuing the same way as in the step above, we can fill the first place in 6 ways, the second place in 5 ways and the sixth place in 1 way. Hence there are 6! ways of forming the words. Once again, we will need to adjust for the two S which can be done by dividing 6! by 2!. Total number of 7 lettered words such that the two Cs are always together = 6!/2! The fraction of seven lettered words such that the two Cs are always together is: = (number of words with two Cs together/total number of words) = (6!/2!)/(7!/[2! x 2!]) = (2!/7) = 2/7 Hence the correct answer choice is A.

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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)42b)210c)420d)840e)5040Correct answer is option 'C'. Can you explain this answer?
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