Here is the detailed solution to the question based on the given conditions:
### Word: DAUGHTER
- Total letters: 8 (D, A, U, G, H, T, E, R)
- Vowels: A, U, E (3 vowels)
- Consonants: D, G, H, T, R (5 consonants)
### (a) The word starts with D and ends with R
Fixing D at the first position and R at the last position, we are left with the letters A, U, G, H, T, E to arrange in the 6 middle positions.
The number of ways to arrange these 6 letters is given by \( 6! \).
Calculation:
6! = 6 × 5 × 4 × 3 × 2 × 1 = 720
Thus, the total number of words = 720.
### (b) Position of letter H remains unchanged
The letter H is fixed at its original position (5th position). This leaves us with 7 positions (D, A, U, G, T, E, R) to arrange the remaining 7 letters. Since the position of H is fixed, we need to arrange the remaining 7 letters in these positions.
The number of ways to arrange the remaining 7 letters is \( 7! \).
Calculation:
7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040
Thus, the total number of words = 5040.
### (c) Relative position of vowels and consonants remains unaltered
The vowels (A, U, E) and consonants (D, G, H, T, R) must maintain their respective original positions in the word DAUGHTER. In this scenario, the three vowels A, U, E can be arranged among themselves, and the five consonants D, G, H, T, R can also be arranged among themselves.
The number of arrangements is:
\( 3! \times 5! \)
Calculation:
3! = 3 × 2 × 1 = 6
5! = 5 × 4 × 3 × 2 × 1 = 120
Total arrangements = 6 × 120 = 720
Thus, the total number of words = 720.
### (d) No two vowels are together
To ensure that no two vowels are together, we first arrange the consonants (D, G, H, T, R) in \( 5! \) ways. This creates 6 gaps (one before the first consonant, four between consonants, and one after the last consonant). We place the three vowels (A, U, E) in any 3 of these 6 gaps, ensuring that no two vowels are adjacent.
The number of ways to choose 3 gaps out of 6 is \( \binom{6}{3} \), and the vowels can be arranged among themselves in \( 3! \) ways.
Calculation:
5! = 120
\( \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \)
3! = 6
Total arrangements = 120 × 20 × 6 = 14400
Thus, the total number of words = 14400.
### (e) All vowels occur together
If all vowels (A, U, E) occur together, we can treat them as a single unit. Along with the consonants D, G, H, T, R, we have 6 units to arrange. The vowels within their single unit can be arranged in \( 3! \) ways.
The number of arrangements is:
\( 6! \times 3! \)
Calculation:
6! = 720
3! = 6
Total arrangements = 720 × 6 = 4320
Thus, the total number of words = 4320.
### (f) All vowels never occur together
If all vowels never occur together, we subtract the cases where all vowels are together from the total number of arrangements of the word.
Total arrangements of DAUGHTER = \( 8! \)
Arrangements where all vowels are together = 4320
The number of arrangements where vowels are not all together is:
\( 8! - 4320 \)
Calculation:
8! = 40320
Total = 40320 - 4320 = 36000
Thus, the total number of words = 36000.
### Final Answers
(a) 720
(b) 5040
(c) 720
(d) 14400
(e) 4320
(f) 36000
