(a) (i) To form a four-digit number using the digits 1, 3, 5, and 6 with no repetition, we can arrange these 4 digits in 4! ways. Thus, the number of different four-digit numbers is \(4! = 24\).

(a) (ii) For odd numbers greater than 500, consider both 3-digit and 4-digit numbers:

3-digit numbers: The hundreds digit can be 5 or 6 (since the number must be greater than 500). The units digit must be odd (1, 3, or 5). Calculate the possibilities:

- 3-digit odd 500+: 4 ways

- 3-digit odd 600+: \(3 \times 2 = 6\) ways

4-digit numbers: The thousands digit can be 1, 3, 5, or 6. The units digit must be odd (1, 3, or 5). Calculate the possibilities:

- 4-digit odd 1000+: 4 ways

- 4-digit odd 3000+: 4 ways

- 4-digit odd 5000+: 4 ways

- 4-digit odd 6000+: 6 ways

Alternatively, consider the arrangement of digits for 4-digit odd numbers:

- Last digit in 3 ways, second to last in 3 ways, second in 2 ways, first in 1 way: 18 ways

Total: 28 ways

(b) Total arrangements of 6 cards is \(6! = 720\). If 4 and 5 are together, treat them as a single unit, so we have 5 units to arrange: \(5! = 120\). Within the unit, 4 and 5 can be arranged in \(2! = 2\) ways. Thus, the number of arrangements where 4 and 5 are together is \(5! \times 2 = 240\).

The number of ways 4 and 5 are not next to each other is \(720 - 240 = 480\).

Probability that 4 and 5 are not next to each other is \(\frac{480}{720} = \frac{2}{3}\).

(i) To form a number between 5000 and 6000, the first digit must be 5. The remaining digits can be chosen from 1, 2, 3, 4, and 6 without repetition. There are 5 choices for the second digit, 4 choices for the third digit, and 3 choices for the fourth digit. Therefore, the total number of numbers is:

\(1 \times 5 \times 4 \times 3 = 60\)

(ii) If repeated digits are allowed, the first digit is still 5. Each of the remaining three digits can be any of the 6 digits (1, 2, 3, 4, 5, 6). Therefore, the total number of numbers is:

\(1 \times 6^3 = 216\)

(i) To find the number of different 6-digit numbers, we use the formula for permutations of multiset:

\(\frac{6!}{3!}\)

where 6 is the total number of digits and 3 is the number of repeated digits (the digit 7 appears three times).

Calculating gives:

\(\frac{6!}{3!} = \frac{720}{6} = 120\)

Thus, there are 120 different 6-digit numbers.

(ii) We need to count the numbers that start and end with an odd digit. The odd digits available are 5 and 7.

Case 1: Start with 5 and end with 7.

We arrange the remaining digits 4, 6, 7, 7. The number of arrangements is:

\(\frac{4!}{2!} = \frac{24}{2} = 12\)

Case 2: Start with 7 and end with 5.

We arrange the remaining digits 4, 6, 7, 7. The number of arrangements is:

\(\frac{4!}{2!} = \frac{24}{2} = 12\)

Case 3: Start with 7 and end with 7.

We arrange the remaining digits 4, 5, 6. The number of arrangements is:

\(4! = 24\)

Adding all cases gives:

\(12 + 12 + 24 = 48\)

Thus, there are 48 numbers that start and end with an odd digit.

(a) The word REFRIGERATOR has 12 letters with repetitions: R appears 3 times, and E appears 2 times. The number of arrangements is given by:

\(\frac{12!}{3! \times 2!}\)

Calculating this gives:

\(\frac{479001600}{6 \times 2} = 9979200\)

(b) If the Rs must all be together, treat the 3 Rs as a single unit. This gives us 10 units to arrange (9 other letters + 1 unit of Rs). The number of arrangements is:

\(\frac{10!}{2!}\)

Calculating this gives:

\(\frac{3628800}{2} = 181440\)

(i) The word 'GOLD MEDAL' consists of 9 letters where D and L are repeated twice each. The number of different arrangements is given by:

\(\frac{9!}{2! \times 2!} = \frac{362880}{4} = 90720\)

(ii) If the two letters D come first and the two letters L come last, we fix these four letters as DD____LL. We need to arrange the remaining 5 letters (G, O, M, E, A) in the middle:

The number of arrangements is given by:

\(5! = 120\)