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There are 11 alphabets A, H, I, M, O, T, U, V, W, X, Y, Z. They are called symmetrical alphabets. The remaining alphabets are known as asymmetrical alphabets.

How many three-lettered words can be formed such that at least one symmetrical letter is there?

  1. $12870$
  2. $18330$
  3. $16420$
  4. None of these
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There are a total of 11 symmetric letters, and therefore, 15 asymmetric letters.

Method 1:  Total number of words possible (no repetition):
                   26*25*24 = 650*24 = 15600

Total number of words possible with only asymmetric letters:
15*14*13 = 210*13 = 2730

Total number of words with at least one symmetric letter:
15600 - 2730 = 12870

Method 2 : 

case1 : Total combination possible with 1 symmetrical and 2 asymmetrical :

The symmetrical number can be in any one of the digits. So, totally 3 possibilities.

Hence, the total combination = 11* 15 * 14 * 3 = 6930

 

case 2: Total combination possible with 2 symmetrical and 1 asymmetrical :

The asymmetrical number can be in any one of the digits. So, totally 3 possibilities.

Hence, the total combination = 11* 10 * 15 * 3 = 4950

 

case 3: Total combination possible with 3 symmetrical:

All the letters symmetrical Hence, the total combination = 11* 10*9 = 990

Hence, total possible  combination= 6930 + 4950 + 990 = 12870.

 

Hence,(A)12870 is the Answer.

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