Q.From a class of 25 students, 10 are to be chosen for an excursion party. There are 3 students who decide that either all of them will join or none of them will join. In how many ways can the excursion party be chosen?

Concept & Intuition

We have 6 distinct letters (S, P, H, E, R, E — note the two E’s are identical? Wait: the word SPHERE has two E’s, so the letters are not all distinct. That’s the key twist. We must treat the multiset: S, P, H, E, E, R. So we are arranging these 6 items into 8 cells, leaving 2 cells empty. The grid has 4 rows, each with 2 cells. The condition “each row contains at least one letter” means no row is completely empty. We count the number of injective assignments of the 6 letters (with identical E’s) to 6 of the 8 cells, then divide by the permutations of the identical E’s. A direct count is messy; inclusion–exclusion over rows is cleaner.

Step-by-step

1. Total arrangements without row restrictions Choose any 6 of the 8 cells to place the letters: $\binom{8}{6}$ ways. Then assign the 6 letters to those chosen cells, accounting for the two identical E’s. The number of distinct sequences of the 6 letters is $\frac{6!}{2!} = 360$. So total unrestricted arrangements:

2. Apply inclusion–exclusion over rows Let $A_i$ be the set of arrangements where row $i$ is empty (no letter in that row). We want arrangements with no empty row:

3. Count $|A_i|$ (one specific row empty) If row $i$ is empty, its 2 cells are forbidden. We have only 6 remaining cells (the other 3 rows). We must place all 6 letters into those 6 cells — so we must use every cell. Number of ways: choose which 6 cells? There’s no choice — they are exactly the 6 cells of the other 3 rows. So just arrange the 6 letters into those 6 cells: $\frac{6!}{2!} = 360$. There are $\binom{4}{1}=4$ such rows, so: $\sum |A_i| = 4 \times 360 = 1440.$ …

Step 1 — Identify the even elements

$A = \{1, 2, 3, \ldots, 10\}$. The even numbers in $A$ are $\{2, 4, 6, 8, 10\}$, which has $5$ elements.

Step 2 — Count subsets containing only even numbers …

Concept and Intuition

We have 6 distinct periods (positions) and 5 distinct subjects (labels). Each period gets exactly one subject, but each subject must appear at least once. That means one subject will appear twice, and the other four subjects appear exactly once. This is a classic “surjective function” counting problem: mapping 6 labeled items onto 5 labeled boxes with no box empty. The direct formula uses Stirling numbers of the second kind and permutations, but we can reason combinatorially.

Step-by-step reasoning

1. Identify the pattern

   Since there are 6 periods and 5 subjects, and no period is vacant, the only way to satisfy “each subject at least once” is that exactly one subject is used twice, and the other four subjects are used once each. So we are choosing which subject gets the extra period.

2. Choose the subject that appears twice

   There are 5 choices for the subject that will occupy two periods.

   Reasoning: This subject will be repeated; the rest appear exactly once.

3. Arrange the 6 periods with this multiset

   After choosing the repeated subject, we have a multiset of 6 items: one subject appears twice, the other four appear once. The number of distinct sequences (assignments of subjects to the 6 ordered periods) is the number of permutations of these 6 items, accounting for the repetition.

   That number is $\frac{6!}{2!} = 360$.

   Why divide by 2!? Because the two identical copies of the repeated subject are indistinguishable in the sequence.

4. Combine the choices …

Concept and Intuition

When a problem asks for “at least one” of something, it’s often easier to count the opposite — numbers that have none of that digit — and subtract from the total. Here, “natural numbers between 100 and 1000” means three-digit numbers from 100 to 999 inclusive. We want those with at least one digit equal to 6. Instead of listing all possibilities (which is messy because the 6 could be in the hundreds, tens, or units place, and could appear more than once), we count all three-digit numbers, then subtract those that avoid the digit 6 entirely.

Step-by-step solution

1. Total three-digit numbers from 100 to 999

   The smallest is 100, the largest is 999.

   Total count = $999 - 100 + 1 = 900$.

2. Count numbers with no digit 6

   For a three-digit number $abc$ (where $a$ is the hundreds digit, $b$ the tens, $c$ the units), we need each digit to be from the set $\{0,1,2,3,4,5,7,8,9\}$ — that’s 9 possible digits, but with a restriction on the hundreds place.

   • Hundreds digit $a$: Cannot be 0 (otherwise it’s not a three-digit number) and cannot be 6. So allowed digits: 1,2,3,4,5,7,8,9 → 8 choices.
   • Tens digit $b$: Can be any digit except 6, including 0. So allowed: 0,1,2,3,4,5,7,8,9 → 9 choices.
   • Units digit $c$: Same as tens → 9 choices.

   Total numbers with no digit 6 = $8 \times 9 \times 9 = 648$. …

Concept & Intuition

This is a counting problem with restrictions (distinctness and a condition on the last digit). The key is to break the selection into independent steps, applying the multiplication principle, but carefully handling the order of choices to avoid overcounting or missing restrictions. A classic pitfall is forgetting that the first digit must also be distinct from the last digit, and that digits come only from 1–9 (no 0). We'll choose the last digit first because it has the special condition (even), then choose the first digit, then the two letters.

Step-by-step solution

1. Choose the last digit (the even digit)

   Digits available: 1, 2, 3, 4, 5, 6, 7, 8, 9.

   Even digits among these: 2, 4, 6, 8 → 4 choices.

   So the last digit can be any of these 4.

2. Choose the first digit

   The first digit must be from 1–9, but distinct from the last digit.

   Since we already used one digit for the last position, we have $9 - 1 = 8$ choices for the first digit.

3. Choose the two letters

   There are 26 English alphabets. The two letters must be distinct.

   • Choose the first letter: 26 choices.
   • Choose the second letter: 25 choices (any letter except the first). Order matters because the code is (letter1)(letter2) — so we do not divide by 2. …

Concept and Intuition

We have 5 distinct positions on a shelf, and we want to fill them with bottles of three varieties (say A, B, C). Each position gets one bottle, and we must use all three varieties at least once. This is a classic "onto" or "surjective" counting problem: the number of functions from a 5-element set (positions) to a 3-element set (varieties) that hit every element of the codomain. The total number of unrestricted assignments is $3^5$, but we must subtract those that miss at least one variety. Inclusion–exclusion handles this cleanly.

Step-by-step solution

1. Total assignments without restriction Each of the 5 positions can be filled with any of the 3 varieties. So the total number of ways is

2. Subtract assignments that miss at least one variety Let $A_1$ be the set of arrangements that miss variety 1 (i.e., use only varieties 2 and 3). Similarly $A_2$ and $A_3$ for missing variety 2 or variety 3.
   • For a specific missing variety, say variety 1, we have only 2 choices per position, so $|A_1| = 2^5 = 32$.
   • There are $\binom{3}{1}=3$ such sets, so the first subtraction gives

3. Add back assignments that miss two varieties If two varieties are missing, only one variety remains. For example, missing varieties 1 and 2 means all bottles are variety 3: exactly 1 way.
   • The number of ways to miss two specific varieties is $1^5 = 1$.
   • There are $\binom{3}{2}=3$ such pairs, so we add back $$147 + 3 \times 1 = 150. …$$

Concept & Intuition

When a problem asks for numbers “greater than” a given number with digit repetition allowed, the natural approach is to count by place value from left to right.

• If the first digit is already larger than the first digit of 4321, then any choice for the remaining three digits works (because the number is already bigger).
• If the first digit equals the first digit of 4321, we must look at the second digit, and so on. This is a classic lexicographic counting method. The pitfall is forgetting that the first digit cannot be 0 (otherwise it’s not a 4‑digit number).

We have digits {0,1,2,3,4,5} and need numbers > 4321.

Step‑by‑step solution

1. Count numbers with first digit > 4

   The first digit can be 5 only (since digits >4 from the set are just 5).

   • Thousands place: 1 choice (5).
   • Hundreds, tens, units: each can be any of the 6 digits (0–5). So count = $1 \times 6 \times 6 \times 6 = 216$.

2. Count numbers with first digit = 4

   Now the number starts with 4, so we need the remaining three digits to make the whole number > 4321.

   We compare digit by digit:

   • Case A: Second digit > 3

     Possible second digits: 4 or 5 (since digits >3 from {0..5} are 4,5).

     For each such choice, the last two digits can be anything (0–5).

     Count = $2 \times 6 \times 6 = 72$.

   • Case B: Second digit = 3

     Now the number starts 43…, so we need the third digit > 2.

     Possible third digits: 3,4,5 (digits >2 from {0..5}).

     For each, the last digit can be anything (0–5).

     Count = $1 \times 3 \times 6 = 18$.

   • Case C: Second digit = 3 and third digit = 2 …

A factor $2^a3^b5^c$ is a perfect square iff $a,b,c$ are all even, with $0\le a\le5$, $0\le b\le6$, $0\le c\le2$:

• $a\in\{0,2,4\}$ — $3$ choices, …

A number divides both $400$ and $1000$ iff it divides $\gcd(400,1000)$.

$400=2^4\cdot5^2,\quad 1000=2^3\cdot5^3\ \Rightarrow\ \gcd=2^3\cdot5^2=200.$ …

We need 4-digit numbers strictly between 1000 and 4000 formed from $\{0,2,3,4\}$ with repetition allowed. …

Let $A$ receive $a$ objects. Then $B=a+1$ and $C=B+2=a+3$. Total: $a+(a+1)+(a+3)=16\Rightarrow3a=12\Rightarrow a=4$, so the split is $A=4$, $B=5$, $C=7$. …

1-digit (1–9): $9$ numbers.

2-digit: first digit $1$–$9$ (9 choices), second digit any of $0$–$9$ except the first (9 choices): $9\times9=81$. …