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Question

In how many distinguishable ways can the letters of the word CHANCE be arranged?

The correct answer is
360

Distinguishable Ways to Arrange Letters

The question asks for the number of unique ways the letters in the word "CHANCE" can be arranged. This is a problem of permutations with repetitions.

Permutation Formula for Repetitions

When arranging items where some items are identical, we use the formula:

$ \text{Number of arrangements} = \frac{n!}{n_1! n_2! \dots n_k!} $

Where:

  • $n$ is the total number of letters.
  • $n_1, n_2, \dots, n_k$ are the frequencies of each distinct repeated letter.

Applying the Formula to "CHANCE"

The word "CHANCE" has 6 letters.

  • Total letters, $n = 6$.
  • The letters are C, H, A, N, C, E.
  • The letter 'C' is repeated 2 times. So, $n_1 = 2$.
  • All other letters (H, A, N, E) appear only once.

Using the formula:

$ \text{Number of arrangements} = \frac{6!}{2!} $

Calculation Steps

  1. Calculate the factorial of the total number of letters: $6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$.
  2. Calculate the factorial of the frequency of the repeated letter: $2! = 2 \times 1 = 2$.
  3. Divide the result from step 1 by the result from step 2: $ \frac{720}{2} = 360 $

Therefore, there are 360 distinguishable ways to arrange the letters of the word CHANCE.

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Important Questions from Permutation and Combination (Notes)

  1. In how many ways can 10 men be divided into two groups of 4 men and 6 men?
  2. In how many ways can you place $N$ coins on a board with $N$ rows and $N$ columns such that every row and every column contains exactly one coin?
  3. The maximum number of points formed by intersection of all pairs of diagonals of convex octagon is
  4. The number of three English letter words, having at least one consonant, but not having two consecutive consonants, is
  5. The number of three digit PINs, in which the third digit is the sum of the first two digits, is
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