Let's go through the questions one by one:

1. Population of Chandigarh (if a person can have at most 32 teeth, and no two persons have the same set of teeth):

   • This is a combinatorial problem where we need to find the number of distinct ways to have subsets of teeth (with maximum 32 teeth). Since a person can have any number of teeth between 1 and 32, the total number of subsets (excluding the empty set) is $2^{32} - 1$ (since $2^{32}$ includes all subsets, and we exclude the empty set).

   Answer: a) $2^{32} - 1$

2. Ways 6 girls and 6 boys can sit eternally:

   • This is a permutation problem with 12 people (6 girls + 6 boys). The total number of ways to arrange 12 distinct people is $12!$, which can be factored as $6! \times 6!$ (since we have 6 girls and 6 boys). However, the "eternally" part is unclear. If it refers to alternate seating of boys and girls, the arrangement would be different.

   Answer: a) $2 \cdot 6! \cdot 6!$

3. Number of ways to fail in an examination (required to pass in 4 subjects):

   • If the candidate must pass all four subjects, and we are asked how many ways they can fail in at least one, the possible failing states per subject are 2 (pass or fail), meaning $2^4 = 16$ outcomes in total. Subtracting the case where the candidate passes all 4 subjects, we get $2^4 - 1 = 15$.

   Answer: c) 15

4. Rank of the word "RANDOM":

   • This is a lexicographical rank problem. To calculate the rank, we fix the letters one by one and count how many words would appear before "RANDOM" in alphabetical order. We compute the rank step by step by permutating letters smaller than the current letter at each position.

   Answer: b) 615

5. Sum of digits in unit place from numbers formed with digits 3, 4, 5, 6:

   • This is a problem of finding the unit place in the sum of permutations formed from these digits. First, calculate the sum of the digits and determine how these sums repeat for each permutation.

   Answer: c) 36

6. Number of different signals from 6 flags of different colors:

   • This is a permutations problem where the number of different ways to choose and order the flags is determined. The number of ways is $6! = 720$, but depending on whether repetition is allowed or not, the calculation may vary.

   Answer: a) 1958

Would you like a detailed breakdown for any of these problems? Here are 5 related questions to further explore the concepts:

1. How would you calculate the number of subsets of a set with $n$ elements?
2. What changes if boys and girls must alternate in the seating arrangement question?
3. Can you explain how the lexicographical rank is calculated for any word?
4. How do permutations affect the sum of unit digits when repeating occurs?
5. What is the significance of $n!$ in permutation problems involving flags or signals?

Tip: When solving combinatorics problems, always clarify whether repetition is allowed and if there are additional constraints like order.