Question

How many different words can be formed using all the letters of the words HONOLULU if no two alike letters are together

Answer 1

720

Answer 2

The problem asks us to find the number of different words that can be formed using all the letters of the word HONOLULU such that no two alike letters are together.

First, let's list the letters and their frequencies in HONOLULU:

• H: 1
• O: 3
• N: 1
• L: 2
• U: 1

Total letters = 1 + 3 + 1 + 2 + 1 = 8.

The condition "no two alike letters are together" means:

1. No two 'O's are adjacent (i.e., 'OO' is not allowed).
2. No two 'L's are adjacent (i.e., 'LL' is not allowed).

We will use the "gap method" or "spacing method" to solve this problem. This method involves arranging the letters that are not restricted first, then placing the restricted letters into the gaps created by the first set of letters. When there are multiple types of restricted letters, the order of placing them matters if the number of gaps changes based on the type of letter placed.

Let's arrange the distinct letters first: H, N, U.

1. Arrange the distinct letters (H, N, U):
   These three letters are all distinct.
   Number of ways to arrange H, N, U = $3! = 3 \times 2 \times 1 = 6$.
   Let's consider one such arrangement, for example: H N U.
   This arrangement creates $3 + 1 = 4$ possible positions (gaps) where other letters can be placed:
   _ H _ N _ U _
2. Place the 'L's (L, L) in the gaps such that no two 'L's are together:
   We have 2 'L's, and they are identical. To ensure no two 'L's are together, we must place them in different gaps.
   There are 4 gaps available, and we need to choose 2 of them for the 'L's.
   Number of ways to choose 2 gaps out of 4 = $^4C_2 = \frac{4 \times 3}{2 \times 1} = 6$.
   For example, if we chose the first and third gaps for H N U, the arrangement would be L H L N U.
   Now, the arrangement consists of 5 letters (H, N, U, L, L), with no two 'L's together.
3. Place the 'O's (O, O, O) in the new gaps such that no two 'O's are together:
   After placing the H, N, U, and L, L letters, we have a total of 5 letters arranged (e.g., L H L N U).
   These 5 letters create $5 + 1 = 6$ possible positions (gaps):
   _ L _ H _ L _ N _ U _
   We have 3 'O's, and they are identical. To ensure no two 'O's are together, we must place them in different gaps.
   There are 6 gaps available, and we need to choose 3 of them for the 'O's.
   Number of ways to choose 3 gaps out of 6 = $^6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$.

The total number of different words formed by multiplying the number of ways at each step:

Total ways = (Ways to arrange H, N, U) $\times$ (Ways to place L's) $\times$ (Ways to place O's)

Total ways = $3! \times ^4C_2 \times ^6C_3 = 6 \times 6 \times 20 = 36 \times 20 = 720$.

The initial method (H,N,U -> L,L -> O,O,O) is correct and gives 720.