Q.The number of possible outcomes when a coin is tossed $6$ times is
(A) $36$
(B) $64$
(C) $12$
(D) $32$

The key idea is that the student is not simply picking one answer; they are choosing a subset of the five alternatives that contains at least two elements. Each alternative can be either selected or not, giving $2^5 = 32$ total subsets. But we must exclude the subsets with 0 or 1 alternative, because the question requires "two or more than two" correct.

Why this works:

Instead of listing all possible selections, we use the fact that the total number of subsets of a set of $n$ items is $2^n$. Then we subtract the forbidden cases (choosing none or exactly one). This is far faster than summing combinations manually.

Step-by-step reasoning:

1. Total possible selections (any number of alternatives):

   For each of the 5 alternatives, the student can either choose it or not. That gives $2^5 = 32$ possible subsets (including the empty set).

2. Identify the forbidden selections:

   The problem says "two or more than two alternatives are correct." So we must exclude:

   • Selecting 0 alternatives: only $\binom{5}{0} = 1$ way.
   • Selecting exactly 1 alternative: $\binom{5}{1} = 5$ ways.

3. Subtract to get the allowed number:

   Allowed ways = $32 - (1 + 5) = 32 - 6 = 26$.

4. Check by direct sum (optional): …

The word FEBRUARY has 8 letters: F, E, B, R, U, A, R, Y. Notice that R appears twice, and the vowels are E, U, A — three distinct vowels. The middle place (second position) must be a vowel. So we first choose which vowel goes in the middle, then fill the first and third positions with any two distinct letters from the remaining 7 letters (since one vowel is used up), but we must account for the fact that the two R's are identical.

Let's work through it step by step.

1. Choose the vowel for the middle position.

   There are 3 vowels: E, U, A. So there are 3 choices for the middle letter.

2. After fixing the middle vowel, we have 7 remaining letters (the original 8 minus the chosen vowel). Among these 7, the letter R appears twice (if the chosen vowel is not R, which it isn't), and all other letters are distinct. So the multiset for the first and third positions is: one repeated letter (R) and 5 other distinct letters.

3. Count the number of ordered pairs (first, third) from these 7 letters.

   We need to count all possible ordered pairs (letter1, letter3) where the two letters can be the same or different, but note that if both are R, that's allowed because there are two R's available. However, we must be careful: the two positions are distinct, so (R, R) is one ordered pair, but it uses both copies of R.

   The total number of ordered pairs without any restriction (allowing repetition) from 7 distinct items would be $7 \times 7 = 49$. But here the 7 items are not all distinct — there are two identical R's. So we cannot simply use $7^2$.

   Better approach: Count directly by cases.

   Case 1: Both first and third are the same letter.

   Which letters can appear twice? Only R, because there are two R's. No other letter appears twice. So the only possibility is (R, R). That's 1 ordered pair.

   Case 2: First and third are different letters.

   We need to choose two distinct letters from the 7 available, but remember that the two R's are identical, so choosing R and some other letter is just one combination (since the two R's are indistinguishable). The number of ways to choose an unordered pair of distinct letters from the multiset:

   • If neither is R: choose 2 from the 5 distinct non-R letters: $\binom{5}{2} = 10$ pairs.
   • If one is R and the other is one of the 5 non-R letters: there are 5 such pairs. So total unordered pairs = $10 + 5 = 15$. …

Concept & Intuition

The divisibility rule for 3 is simple: a number is divisible by 3 exactly when the sum of its digits is divisible by 3.

Here we are forming 3-digit numbers from the digits 1, 3, 5, 7, 9 (all odd, all distinct).

We don’t need to list every number — we just need to find all sets of three digits whose sum is a multiple of 3, then count how many distinct 3-digit numbers each set can form.

Step-by-step reasoning

1. List the digits and their remainders modulo 3

   • 1 → remainder 1
   • 3 → remainder 0
   • 5 → remainder 2
   • 7 → remainder 1
   • 9 → remainder 0

   So we have:

   • Remainder 0: {3, 9}
   • Remainder 1: {1, 7}
   • Remainder 2: {5}

2. When does the sum of three digits become a multiple of 3?

   The sum of three numbers is divisible by 3 if the sum of their remainders (mod 3) is 0.

   Possible remainder combinations (order doesn’t matter) that sum to 0 mod 3:

   • (0,0,0)
   • (1,1,1)
   • (2,2,2)
   • (0,1,2)

3. Check which combinations are possible with our digits

   • (0,0,0): need three digits from {3,9} — only two available. Impossible.
   • (1,1,1): need three digits from {1,7} — only two available. Impossible.
   • (2,2,2): need three digits from {5} — only one available. Impossible.
   • (0,1,2): we have exactly two digits with remainder 0, two with remainder 1, and one with remainder 2. So we can pick one from each group.

4. List all triples of digits that work

   Choose one from remainder 0: {3,9} → 2 choices

   Choose one from remainder 1: {1,7} → 2 choices …