Number of ways of attempting 1^st, 2^nd, 3^rd question are each.
Total number of ways
4³ = 4 × 4 × 4 = 64
Number of ways, getting correct answers = 1³ = 1
∴ Number of ways of not getting all answer correct = 64 - 1 = 63

APPLE = 5 letters.
But two letters PP is of same kind.
Thus, required permutations,
$$\eqalign{ & = \frac{{5!}}{{2!}} \cr & = \frac{{120}}{2} \cr & = 60 \cr} $$

There are 8 students and maximum capacity of cars toger is 9.
We may divide 8 students as follows:

Case I:
5 students in first car and 3 in second.
Hence, 8 students are divided into groups of 5 and 3 in ⁸C₃ = 56 ways.

Case II:
4 students in first car and 4 in second.
So, 8 students are divided into two groups of 4 and 4 in ⁸C₄ = 78 ways.
Therefore, total number of ways in which 8 students can travel is:
56 + 70 = 126

At least 1 question from each section is compulsory, so from 1^st section candidate can attempt 1 or 2 or 3 or 4 questions.
In each section each question can be dealt with in 2 ways, i.e. eir he attempts it or leaves it.
So far 4 question re are 2 × 2 × 2 × 2 ways to attempt.
As he has to attempt at least 1 question, total number of ways in which he can attempt questions from 1st section is 2⁴ - 1
Similarly for 2^nd section re are 2⁵ - 1 ways in which he can attempt and for 3^rd section re are 2⁶ - 1 ways.
The ways in which attempts one or more questions in any section is independent of number of ways in which he attempts one or more questions from or sections.
Thus, total number of ways in which he can attempt questions in that paper:
= (2⁴ - 1)(2⁵ - 1)(2⁶ - 1)
= 15 × 31 × 63
= 29295

¹⁰C₅ : when both are not included.
¹⁰C₄ : when one of m is included.

Number of ways
= ¹⁰C₅ + ¹⁰C₄ + ¹⁰C₄
= 672

Out of 26 alphabets two distinct letters can be chosen in ²⁶P₂ ways.
Coming to numbers part, re are 10 ways (any number from 0 to 9 can be chosen) to choose first digit and similarly anor 10 ways to choose second digit.
Hence, re are totally 10 × 10 = 100 ways.
Combined with letters re are,
²⁶P₂ × 100 = 65000 ways to choose vehicle numbers.

Number of ways a pair of shoes can be selected,
= ³C₁
= 3 ways
Number of ways lower wear can be selected = (3 + 4) = 7 ways.
Number of ways upper wear can be selected, = 3 + 6 + (3 × 6)
= 27 ways
Number of ways jacket can be chosen or not chosen = 3 ways {No jacket, 1^st jacket, 2^nd jacket}.
Total number of different outfits
= 3 × 7 × 27 × 3
= 1701 ways

There are 9 non-zero digits to arrange mselves at 4 different position. Each letter can be arrange at different position in 9 different ways.
So, required number of ways,
= 9 × 9 × 9 × 9
= 9⁴

The given words contains 6 letters of which A is taken 3 times, N is taken 2 times and rest are all different.
∴ Required number of ways
$$\eqalign{ & = \frac{{6!}}{{3! \times 2!}} \cr & = \frac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{6 \times 2}} \cr & = 60 \cr} $$

Number of words which have at least one letter replaced,
= Total number of words - total number of words in which no letter is repeated.
= 10⁵ – ¹⁶P₅
= 100000 − 30240
= 69760