Step 1: Count the total number of letters in the word "DEPENDENT". The word has 9 letters: D, E, P, E, N, D, E, N, T.
Step 2: Count the frequency of each distinct letter. - D appears 2 times - E appears 3 times - P appears 1 time - N appears 2 times - T appears 1 time Total letters \( n = 9 \).
Step 3: Apply the formula for permutations with repetitions. The number of distinct permutations of \( n \) objects where there are \( n_1 \) identical objects of type 1, \( n_2 \) identical objects of type 2, ..., \( n_k \) identical objects of type k is given by: \[ \frac{n!}{n_1! n_2! \cdots n_k!} \] In this case, \( n = 9 \), and the counts of repeated letters are \( n_D = 2 \), \( n_E = 3 \), \( n_N = 2 \).
Step 4: Substitute the values into the formula. \[ \text{Number of permutations} = \frac{9!}{2! \times 3! \times 2!} \]
Step 5: Calculate the factorial values and the final result. \[ 9! = 362,880 \] \[ 2! = 2 \] \[ 3! = 6 \] \[ 2! = 2 \] \[ \text{Number of permutations} = \frac{362,880}{2 \times 6 \times 2} = \frac{362,880}{24} \] \[ \text{Number of permutations} = 15,120 \]
Step 6: Compare the result with the given options. The calculated number of permutations is 15,120, which matches option (A).
If all the words with or without meaning made using all the letters of the word "KANPUR" are arranged as in a dictionary, then the word at 440th position in this arrangement is:
The number of 6-letter words, with or without meaning, that can be formed using the letters of the word MATHS such that any letter that appears in the word must appear at least twice, is $ 4 \_\_\_\_\_$.
Which of the following is an octal number equal to decimal number \((896)_{10}\)?