Question

A classic counting problem is to determine the number of different ways that the letters of "occasionally" can be arranged. find that number.

Answer 1

The word has 12 letters in total, distributed as follows:

2 o's
2 c's
2 a's
1 s
1 i
1 n
2 l's
1 y.

There are $12!=12\cdot 11\cdot ... \cdot 3\cdot 2$ permutations of 12 objects. Nevertheless, since we have repeating letters, swapping them would still return the same words.

So, the new total is

$\frac{12!}{2!2!2!2!} = 29937600$

Answer 2

Answer with explanation:

⇒Number of Alphabets in the word " OCCASIONALLY"

                              = 12

⇒Combination of Alphabets in  word " OCCASIONALLY"

     =  O---2, C--2, A--2,S---1, I---1, N---1, L---2, Y----1

⇒Number of ways by which Letters of Word  " OCCASIONALLY" can be Arranged, as order of Alphabets is also Important,So we will use the concept of Permutation.Also, letter O,C, A and L appear 2 times .So we will divide it to total number of Alphabets taking factorial of each number as well as total number of Alphabets.

     $=\frac{\text{Total number of Alphabets in the word OCCASIONALY}}{2! \times 2! \times 2! \times 2! }\\\\=\frac{\text{12!}}{2! \times 2! \times 2! \times 2! }\\\\=\frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{16}\\\\=12 \times 11 \times 10 \times 9 \times 7 \times 6 \times 5 \times 4 \times 3 \times 1\\\\=29937600$