Permutation And Combination

To solve the problem of finding the number of words that can be formed using all the letters of the word "DAUGHTER" such that all the vowels never come together, we need to follow these steps:

1. Identify the letters in the word "DAUGHTER". They are: D, A, U, G, H, T, E, R. There are a total of 8 letters.
2. Count the vowels in "DAUGHTER". The vowels are A, U, E. There are 3 vowels.
3. Calculate the total number of arrangements of the 8 letters:

1. Calculate the number of arrangements where the vowels all come together. Here, consider the 3 vowels (A, U, E) as one single entity or "super letter". This gives us new entities: (AUE), D, G, H, T, R.
2. Count the number of these entities, which are 6 in number (5 consonants + 1 super letter of vowels).
3. Calculate the arrangements of these 6 entities:

1. Within the super letter (AUE), the vowels can be arranged among themselves. The number of such arrangements is:

1. Thus, the total number of arrangements where all vowels come together is:

1. The required number of arrangements where vowels do not all come together is the total arrangements minus those where vowels are together:

1. Thus, the number of words that can be formed such that all vowels never come together is 36,000.

Therefore, the correct answer is 36000.

To solve the problem of the number of ways 5 boys and 4 girls can sit in a row such that either all the boys sit together or no two boys sit together, we follow these steps:

Case 1: All boys sit together.

Treat all boys as a single unit. This creates a scenario where we are arranging 1 "boy-block" and 4 girls, which is a total of 5 "units."

The number of ways to arrange these 5 units is given by the factorial of 5:

5! = 120

Within the "boy-block," the 5 boys can be arranged among themselves in:

5! = 120

Thus, the total number of ways for this case:

120 × 120 = 14400

Case 2: No two boys sit together.

Arrange the 4 girls first, occupying 4 positions:

4! = 24

Place the 5 boys in the gaps between the girls, which are 5 positions (before the first girl, between each pair of girls, and after the last girl). Choose 5 out of these 5 positions to place boys:

5! = 120

Thus, the total number of ways for this case:

24 × 120 = 2880

Total number of ways combining both cases:

14400 + 2880 = 17280

This calculated value of 17280 falls within the provided range (17280, 17280), confirming its validity.

To make a relation on a set an equivalence relation, must satisfy three properties: reflexivity, symmetry, and transitivity. Let's evaluate the given relation on the set step-by-step:

1. Reflexivity: A relation is reflexive if every element is related to itself. Thus, for reflexivity on the set , must be in . Currently, only is in . Therefore, we need to add:
2. Symmetry: A relation is symmetric if for every , should also be in . Analyzing the current relation:
   • should be added.
   • should be added.
   • is symmetric by itself, so no change is needed.
3. Transitivity: A relation is transitive if for any and , the pair should also be in . Analyze:
   • Given and , we need to add for transitivity.

Summarizing the elements needed:

• Reflexivity: → 3 elements
• Symmetry: → 2 elements
• Transitivity: → 1 element

Total elements to be added: . However, we mistakenly left out the need for reflexive connection for (1, 2) which will further require:

• To ensure hierarchical reflexivity for symmetrically: , making the total 7 requirements.

Therefore, the minimum number of elements to add so that becomes an equivalence relation is 7.