(i) The word ARGENTINA has 9 letters, with the letter 'A' repeated twice and the letter 'N' repeated twice. The number of different arrangements is given by:

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

(ii) For arrangements with a consonant at the beginning, then a vowel, and so on alternately, we have 5 positions for consonants and 4 positions for vowels. The number of ways to arrange the consonants is \(5!\) and the number of ways to arrange the vowels is \(4!\). Therefore, the total number of such arrangements is:

\(\frac{5! \times 4!}{2!2!} = \frac{120 \times 24}{4} = 720\)

(i) The number 1223678 has 7 digits, with the digit '2' repeated twice. The number of different permutations is given by:

\(\frac{7!}{2!} = \frac{5040}{2} = 2520\)

(ii) Consider the digits 1, 3, 7 as a single block. This block, along with the other 4 digits (2, 2, 6, 8), forms 5 blocks in total. The number of permutations of these 5 blocks is:

\(\frac{5!}{2!} = \frac{120}{2} = 60\)

Within the block of 1, 3, 7, the digits can be arranged in \(3!\) ways. Therefore, the total number of permutations is:

\(60 \times 6 = 360\)

(iii) For the number to be even, it must end in 2, 6, or 8. Calculate the permutations for each case:

Ending in 2: Consider the remaining digits 1, 2, 3, 6, 7. The number of permutations is:

\(\frac{6!}{2!} = \frac{720}{2} = 360\)

Ending in 6 or 8: The number of permutations for each is:

\(\frac{6!}{2!} = 360\)

Summing these gives:

\(360 + 360 + 360 = 1440\)

(a) Treat the two As as a single unit and the two Ls as a single unit. This gives us 7 units to arrange: AA, LL, I, G, A, T, O, R. The number of arrangements is given by:

\(7! = 5040\)

(b) First, calculate the total number of arrangements of the 9 letters:

\(\frac{9!}{2!2!} = 90720\)

Now, calculate the number of favorable arrangements where the two Ls are together and there are exactly 6 letters between the two As. Consider the arrangement of 6 letters including Ls between As:

\(5! \times 5 \times 2 = 1200\)

The probability is then:

\(\frac{1200}{90720} = 0.0132\)

(a) The word CROCODILE consists of 9 letters where C and O are repeated. The formula for permutations of a multiset is given by:

\(\frac{9!}{2!2!}\)

Calculating this gives:

\(\frac{362880}{4} = 90,720\)

(b) To find the number of arrangements with C at each end and the two Os not together, we use the following method:

First, fix C at both ends, leaving 7 positions to fill with the letters R, O, C, O, D, I, L.

Calculate the total arrangements of these 7 letters:

\(\frac{7!}{2!} = 2520\)

Now, calculate the arrangements where the two Os are together. Treat the two Os as a single unit, so we have 6 units to arrange: (OO), R, C, D, I, L.

Calculate the arrangements of these 6 units:

\(6! = 720\)

Subtract the arrangements where Os are together from the total arrangements:

\(2520 - 720 = 1800\)

(a) Consider the three Es as a single unit and the two Ds as another single unit. This reduces the problem to arranging the units: {EEE, DD, C, I, V}. There are 5 units to arrange, so the number of arrangements is given by:

\(5! = 120\)

(b) First, find the total number of arrangements of the letters in DECEIVED. The word has 8 letters with repetitions of E and D. The total number of arrangements is:

\(\frac{8!}{2!3!} = 3360\)

Next, find the number of arrangements where all three Es are together. Treat the three Es as a single unit, reducing the problem to arranging {EEE, D, D, C, I, V}. The number of arrangements is:

\(\frac{6!}{2!} = 360\)

Therefore, the number of arrangements where the three Es are not all together is:

\(3360 - 360 = 3000\)