All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)

Required Probability $= \frac{26 \times 25 \times 24 \times 9 \times 8}{26 \times 26 \times 26 \times 9 \times 9}$

(ii)

Required Probability $= \frac{1-\mathrm{P}(\mathrm{same digits})}{2}$
$= \frac{1-\frac{9}{9^{2}}}{2}$
$= \frac{4}{9}$

(iii)
Case 1: 2 same letter and 2 different digits

$26 \times 25 \times \frac{3!}{2!} \times {^9\!P_2}=140400$

Case 2: different letter and 2 same digits

$9 \times {^26\!P_3}=140400$

Case 3:3 same letter and 2 same digits

$9 \times 26 =234$

Required Probability $= \frac{140400+140400+234}{26 \times 26 \times 26 \times 9 \times 9} = 0.197$

(iv)
Case 1: 2 different consonants, 1 vowel & 1 even digit

${^{21}\!C_2} \times {^5\!C_1} \times 3! \times {^4\!C_1} \times {^5\!C_1} \times 2! = 252000$

Case 2: 2 same consonant, 1 vowel & 1 even digit

${^{21}\!C_1} \times {^5\!C_1} \times \frac{3!}{2!} \times {^4\!C_1} \times {^5\!C_1} \times 2! = 12600$

Required Probability $= \frac{252000+12600}{26 \times 26 \times 26 \times 9 \times 9} = 0.186$

This question is not that easy. A lot of thinking is required; but that is the case with all permutation questions. I strongly suggest students to not spend too much time on permutation questions and come back to solve it later.

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