Counting

How many different ways can 4 students be seated in 4 chairs?

Solution: Since all 4 students are seated, the number of ways = \( P(4, 4) = 4! = 24 \)

In how many ways can 3 students be selected and arranged from a group of 6?

Solution: \( P(6, 3) = \frac{6!}{(6-3)!} = \frac{720}{6} = 120 \)

How many 3-digit numbers can be formed using the digits 1, 2, 3, 4, 5 with no repetition?

Solution: Choose and arrange 3 digits from 5: \( P(5, 3) = \frac{5!}{2!} = 60 \)

How many ways can 4 books be arranged on a shelf if one specific book must always be at the end?

Solution: Fix the specific book at the end. Now arrange the remaining 3 books: \( 3! = 6 \)

In how many ways can 5 people be seated around a circular table?

Solution: Circular permutations: \( (n - 1)! = 4! = 24 \)

How many permutations can be formed from the letters of the word LEVEL?

Solution: Total letters = 5; Repeated letters: L (2), E (2)

\[ \text{Permutations} = \frac{5!}{2! \cdot 2!} = \frac{120}{4} = 30 \]

From 7 people, how many ways can a president, vice president, and secretary be chosen?

Solution: These are distinct positions; order matters: \( P(7, 3) = \frac{7!}{4!} = 210 \)

In how many ways can the letters of the word SCHOOL be arranged such that S and C are together?

Solution: Treat SC as one unit. So we have: [SC], H, O, O, L → 5 letters

But O is repeated.

\[ \text{Arrangements} = \frac{5!}{2!} = \frac{120}{2} = 60 \]

Also SC and CS both allowed → multiply by 2: \( 60 \times 2 = 120 \)

How many 4-digit numbers can be formed using the digits 1 to 6, such that the number starts with an odd digit?

Solution: Odd digits: 1, 3, 5 → 3 choices for first digit

Remaining 3 digits from 5 unused digits: \( P(5, 3) = 60 \)

Total = \( 3 \times 60 = 180 \)

How many ways can the letters of the word APPLE be arranged if P’s must be together?

Solution: Treat PP as a block → Remaining: [PP], A, L, E → 4 elements

\[ \text{Arrangements} = 4! = 24 \]

How many 5-digit numbers can be formed using digits 1–5 such that no digit repeats and the number is even?

Solution: For a number to be even, it must end in 2 or 4.

Case 1: Last digit is 2 → Remaining 4 digits from {1,3,4,5}: \(4!\)

Case 2: Last digit is 4 → Remaining digits {1,2,3,5}: \(4!\)

\[ \text{Total} = 2 \times 4! = 2 \times 24 = \boxed{48} \]

How many ways can the letters of the word ENGINEER be arranged so that all vowels are together?

Solution: Vowels = E, I, E, E → 4 vowels (E repeated 3 times) Consonants = N, G, N, R → 4 consonants (N repeated 2 times)

Treat vowels as one block → total blocks = 5

\[ \text{Ways to arrange blocks} = \frac{5!}{2!} = 60 \] \[ \text{Vowel arrangements} = \frac{4!}{3!} = 4 \]

\[ \text{Total} = 60 \times 4 = \boxed{240} \]

How many 4-digit numbers can be formed using digits 0-9 such that no digit repeats and at least one digit is 7?

Solution: Total 4-digit numbers with distinct digits:

- Thousands digit: 1–9 → 9 options - Remaining 3 digits from remaining 9 digits → \( P(9,3) \)

\[ \text{Total} = 9 \times P(9,3) = 9 \times 504 = 4536 \]

Now, subtract those without 7:

- Choose 4 digits from {0–9} excluding 7 → 9 digits - Thousands place $\neq$ 0 $\implies$ valid starting digits = 8 options (excluding 0 and 7) - Remaining 3 digits: choose from remaining 8

\[ \text{Without 7} = 8 \times P(8,3) = 8 \times 336 = 2688 \]

\[ \text{With at least one 7} = 4536 - 2688 = \boxed{1848} \]

In how many ways can 6 people be arranged in a row if 2 particular people (A and B) must always occupy the first and last positions (in any order)?

Solution: A and B can be in (1st,6th) or (6th,1st) → 2 ways Remaining 4 people can be arranged in \(4!\)

\[ \text{Total} = 2 \times 4! = 2 \times 24 = \boxed{48} \]

How many arrangements of the word ASSASSIN are there such that no two S’s are adjacent?

Solution: Total letters = 8 S appears 4 times.

First, arrange the non-S letters: A, A, I, N → 4 letters Arrangements = \( \frac{4!}{2!} = 12 \)

Place 4 S's in the 5 available gaps between/around non-S's: _ A _ A _ I _ N _ → 5 positions

Choose 4 positions out of 5 for S’s: \( \binom{5}{4} = 5 \)

\[ \text{Total} = 12 \times 5 = \boxed{60} \]

How many 3-digit numbers can be formed using digits 2, 3, 4, 5, 6, 7 such that 5 is always used and digits are not repeated?

Solution: Total 3-digit numbers using digits from set of 6.

We fix 5 in one position. Then choose 2 more from remaining 5 digits.

Case 1: 5 at hundreds place → choose 2 digits from 5 → \(P(5,2) = 20\)

Case 2: 5 at tens place → hundreds digit $\neq$ 0, 5 → choose from 5 digits (excluding 5), then one more from remaining 4 → \(5 \times 4 = 20\)

Case 3: 5 at units place → similar → \(5 \times 4 = 20\)

\[ \text{Total} = 20 + 20 + 20 = \boxed{60} \]

In how many distinct ways can 8 different people be seated around a circular table if persons \(A\) and \(B\) have exactly two people sitting between them?

Solution: Fix \(A\)'s position to remove rotational symmetry. Around the circle, \(B\) must be placed so that exactly two seats lie between \(A\) and \(B\). There are \(2\) such positions for \(B\) (clockwise or counterclockwise).

Arrange the remaining \(6\) people in the remaining seats: \(6!\) ways.

\[ \text{Total} = 2 \times 6! = 2 \times 720 = \boxed{1440}. \]

In how many ways can 6 people be arranged in a row if person $A$ must be in one of the two end positions?

Solution: First, choose the position for $A$: There are \(2\) choices (left end or right end).

The remaining \(5\) people can be arranged in any order in the remaining \(5\) seats: \(5! = 120\) ways.

\[ \text{Total arrangements} = 2 \times 120 = \boxed{240}. \]

How many ways can the letters of EXAMPLE be arranged such that no two vowels are adjacent?

Solution: Vowels = E, A, E → total = 3 (E repeated twice)

Consonants = X, M, P, L → 4 letters

Arrange consonants: \(4! = 24\)

Now place 3 vowels in gaps between consonants: _ C _ C _ C _ C _ → 5 gaps

Choose 3 of 5 gaps: \( \binom{5}{3} = 10 \)

Vowel arrangements = \( \frac{3!}{2!} = 3 \)

\[ \text{Total} = 24 \times 10 \times 3 = \boxed{720} \]

How many ways can 5 boys and 3 girls be seated in a row such that no two girls sit together?

Solution: First, place 5 boys: \(5! = 120\)

Gaps between boys = 6 positions: _ B _ B _ B _ B _ B _

Choose 3 gaps from 6 to place girls: \( \binom{6}{3} = 20 \)

Arrange girls in those positions: \(3! = 6\)

\[ \text{Total} = 120 \times 20 \times 6 = \boxed{14400} \]

In how many ways can the letters of the word HUNTER be arranged such that the word does not start with H?

Solution: Total arrangements = \(6! = 720\)

Fix H at first position: Remaining 5 letters → \(5! = 120\)

\[ \text{Valid arrangements} = 720 - 120 = \boxed{600} \]

How many 4-digit numbers greater than 3000 can be formed using digits 2, 3, 4, 5, 6 without repetition?

Solution: Thousands digit must be 3, 4, 5, or 6 (4 options) Then choose 3 more digits from remaining 4 → \(P(4,3) = 24\)

\[ \text{Total} = 4 \times 24 = \boxed{96} \]

In how many ways can the letters of the word COMMITTEE be arranged such that all T’s are together?

Solution: Letters = 9; Repeats: M(2), T(2), E(2)

Treat TT as a block → 8 items: C, O, M, M, I, TT, E, E

\[ \text{Total} = \frac{8!}{2! \cdot 2! \cdot 1!} = \frac{40320}{4} = \boxed{10080} \]

How many 4-digit numbers can be formed from digits 1 to 9 such that digit 5 appears at least once?

Solution: Total 4-digit numbers using 1–9 (no 0): \(9 \times 8 \times 7 \times 6 = 3024\)

Exclude numbers without 5:

- Choose 4 digits from 8 (excluding 5): \(P(8,4) = 1680\)

\[ \text{With at least one 5} = 3024 - 1680 = \boxed{1344} \]

In how many ways can 5 boys and 3 girls be seated in a row such that all the girls sit together?

Solution: Treat the 3 girls as a block → now 6 "people" to arrange → \(6! = 720\)

Girls within their block can be arranged in \(3! = 6\)

\[ \text{Total} = 720 \times 6 = \boxed{4320} \]

How many arrangements of the word BALLOON can be made such that both L’s are not together?

Solution: Total arrangements without restriction: Letters = 7; L(2), O(2)

\[ \text{Total} = \frac{7!}{2! \cdot 2!} = \frac{5040}{4} = 1260 \]

Now count cases where L’s are together:

Treat LL as a block → total = 6 items: B, A, LL, O, O, N O(2): \[ \text{With L’s together} = \frac{6!}{2!} = 360 \]

\[ \text{Not together} = 1260 - 360 = \boxed{900} \]

How many 3-digit even numbers can be formed using digits 2, 3, 4, 5, 6 with no digit repeated?

Solution: Even numbers end in 2, 4, or 6 → 3 cases

Case 1: End with 2 → choose 2 digits from {3,4,5,6} → \(P(4,2) = 12\)

Case 2: End with 4 → {2,3,5,6} → \(P(4,2) = 12\)

Case 3: End with 6 → {2,3,4,5} → \(P(4,2) = 12\)

\[ \text{Total} = 3 \times 12 = \boxed{36} \]

From the word STATISTICS, how many distinct arrangements are there where all S’s are together?

Solution: Letters: 10; S(3), T(3), I(2), A(1), C(1)

Group all S’s → treat as a block: 8 items: SSS, T(3), I(2), A, C

\[ \text{Total} = \frac{8!}{3! \cdot 2!} = \frac{40320}{12} = \boxed{3360} \]

How many ways can 6 people be seated in a row such that A and B have exactly one person between them?

Easy Solution (Position Method): Valid seat pairs for A and B with exactly one seat between them are: \((1,3),(2,4),(3,5),(4,6)\) — \(\;4\) pairs.

For each pair: - A and B can swap places: \(\;2\) ways. - Choose who sits between them: from the remaining \(4\) people: \(\;4\) ways. - Arrange the remaining \(3\) people in the remaining \(3\) seats: \(\;3! = 6\) ways.

\[ \text{Total} = 4 \times 2 \times 4 \times 3! = 4 \times 2 \times 4 \times 6 = \boxed{192}. \]

(Quick Check) Alternatively, fix a valid pair (e.g., \(1,3\)) and order A/B in \(2\) ways; then the other \(4\) people fill the remaining seats in \(4! = 24\) ways. So \(4 \times 2 \times 24 = \boxed{192}\).

In how many ways can 5 identical red balls and 4 identical green balls be arranged in a row?

Solution: Total objects = 9 Red balls identical, Green balls identical

\[ \text{Total arrangements} = \frac{9!}{5! \cdot 4!} = \frac{362880}{120 \cdot 24} = \boxed{126} \]

Use combinations when:

• The selection is made without regard to order.

• Groupings or committees are formed.

• No specific arrangement or position is mentioned.

Avoid using permutations unless the sequence/arrangement matters.

From a group of 10 people, how many ways can we select 4 to form a committee?

Solution: \[ \binom{10}{4} = \frac{10!}{4! \cdot 6!} = \frac{5040}{24} = \boxed{210} \]

In a party of 12 people, if each person shakes hands with every other person once, how many handshakes occur?

Solution: Each handshake is a combination of 2 people: \[ \binom{12}{2} = \frac{12 \cdot 11}{2} = \boxed{66} \]

From a bag of 15 balls, how many ways can 6 balls be drawn simultaneously?

Solution: Order doesn't matter: \[ \binom{15}{6} = \boxed{5005} \]

A class has 8 boys and 7 girls. In how many ways can a team of 5 students be chosen such that it includes at least 2 girls?

Solution: We sum combinations for cases:

• 2 girls, 3 boys: \(\binom{7}{2} \cdot \binom{8}{3} = 21 \cdot 56 = 1176\)

• 3 girls, 2 boys: \(\binom{7}{3} \cdot \binom{8}{2} = 35 \cdot 28 = 980\)

• 4 girls, 1 boy: \(\binom{7}{4} \cdot \binom{8}{1} = 35 \cdot 8 = 280\)

• 5 girls: \(\binom{7}{5} = 21\)

Total = \(1176 + 980 + 280 + 21 = \boxed{2457}\)

How many 5-card hands can be drawn from a deck of 52 cards?

Solution: \[ \binom{52}{5} = \boxed{2,598,960} \]

In how many ways can 3 identical prizes be given to 10 students?

Solution: Choose 3 students from 10: \[ \binom{10}{3} = \boxed{120} \]

Out of 20 questions in an exam, a student must attempt 15. In how many ways can this be done?

Solution: \[ \binom{20}{15} = \binom{20}{5} = \boxed{15504} \]

How many subsets of 4 elements can be formed from a set of 10 elements?

Solution: \[ \binom{10}{4} = \boxed{210} \]

A lottery draws 6 numbers out of 49. How many combinations are possible?

Solution: \[ \binom{49}{6} = \boxed{13,983,816} \]

From 12 different books, in how many ways can 5 books be selected for reading?

Solution: \[ \binom{12}{5} = \boxed{792} \]

In how many ways can a committee of 4 people be selected from 7 men and 5 women such that the committee contains at least 2 women?

Solution:

We consider all valid combinations: \[ \begin{aligned} &\text{2 women, 2 men: } \binom{5}{2} \cdot \binom{7}{2} = 10 \cdot 21 = 210 \\ &\text{3 women, 1 man: } \binom{5}{3} \cdot \binom{7}{1} = 10 \cdot 7 = 70 \\ &\text{4 women: } \binom{5}{4} = 5 \end{aligned} \]

\[ \text{Total} = 210 + 70 + 5 = \boxed{285} \]

From a group of 10 people, in how many ways can a team of 5 be formed such that two particular persons are never together?

Solution:

Let the two particular persons be A and B.

Total combinations without any restriction: \(\binom{10}{5} = 252\)

Number of teams where A and B are together: Choose A and B (fixed), then choose 3 more from remaining 8: \[ \binom{8}{3} = 56 \]

Required: \(252 - 56 = \boxed{196}\)

Question: For a game in which two partners oppose two other partners, a total of \(6\) men are available. If every possible pair must play with every other pair, how many games are played?

Solution:

We interpret a "game" as selecting two disjoint unordered pairs (pair A vs pair B). Each distinct unordered pair-of-pairs corresponds to one game.

Method 1 (choose pairs stepwise):

1. Choose the first pair of players: there are \[ \frac{6\times5}{2\times1}=15 \] ways to pick 2 out of 6.

2. From the remaining 4 men choose the opposing pair: \[ \frac{4\times3}{2\times1}=6 \] ways.

3. The ordered choice (first pair, then second pair) counts each matchup twice (pair \(P\) vs \(Q\) and pair \(Q\) vs \(P\) are the same game), so divide by 2: \[ \frac{15\times 6}{2} = 45. \]

Method 2 (choose 4 players then partition):

1. Choose which 4 of the 6 men will play in a particular game: \[ \frac{6\times5\times4\times3}{4\times3\times2\times1}=15 \] ways.

2. For those 4 chosen players, count the number of ways to split them into two unordered pairs. The pairings of 4 distinct people into two unordered pairs equals \(3\) (you can check: fix one person, they have 3 choices for partner, then remaining 2 form the other pair; divide by 2 to remove order of the two pairs: \(\tfrac{3}{1}=3\)).

3. Multiply: \[ 15 \times 3 = 45. \]

\[ \boxed{45} \]

Answer: \(\boxed{45}\) games.

A question paper has 10 questions. A student has to answer exactly 7. However, questions 1 and 2 are mutually exclusive (i.e., both cannot be selected together). In how many ways can the student choose 7 questions?

Solution:

Case 1: Choose Q1 but not Q2 → choose 6 from remaining 8: \(\binom{8}{6} = 28\)

Case 2: Choose Q2 but not Q1 → again, \(\binom{8}{6} = 28\)

Case 3: Choose neither Q1 nor Q2 → choose 7 from remaining 8: \(\binom{8}{7} = 8\)

Total = \(28 + 28 + 8 = \boxed{64}\)

How many ways can you choose a group of 6 people from 12 such that two particular people are either both in or both out of the group?

Solution:

Let A and B be the two particular people.

Case 1: Both A and B are in: choose 4 more from 10 → \(\binom{10}{4} = 210\)

Case 2: Both A and B are out: choose 6 from 10 → \(\binom{10}{6} = 210\)

Total = \(210 + 210 = \boxed{420}\)

From 8 different books, how many ways can 5 be chosen such that at least one specific book is included?

Solution:

Total ways without restriction: \(\binom{8}{5} = 56\)

Ways that exclude the specific book: \(\binom{7}{5} = 21\)

Required: \(56 - 21 = \boxed{35}\)

From 9 items, including 3 identical ones, how many distinct sets of 4 items can be chosen?

Solution:

We count based on how many of the identical items are included.

Case 1: 0 identical → choose 4 from 6 distinct: \(\binom{6}{4} = 15\)

Case 2: 1 identical → \(\binom{6}{3} = 20\)

Case 3: 2 identical → \(\binom{6}{2} = 15\)

Case 4: 3 identical → \(\binom{6}{1} = 6\)

Total = \(15 + 20 + 15 + 6 = \boxed{56}\)

How many ways can a team of 4 be formed from 6 girls and 4 boys such that the team includes more girls than boys?

Solution:

Valid compositions: - 3 girls, 1 boy → \(\binom{6}{3} \cdot \binom{4}{1} = 20 \cdot 4 = 80\) - 4 girls → \(\binom{6}{4} = 15\)

Total = \(80 + 15 = \boxed{95}\)

From 12 students, how many ways can 5 be chosen such that two specific students are not both included?

Solution:

Total combinations: \(\binom{12}{5} = 792\)

Combinations with both A and B: Choose 3 more from 10 → \(\binom{10}{3} = 120\)

Required: \(792 - 120 = \boxed{672}\)

From 6 red, 5 green, and 4 blue balls, how many ways can 5 balls be selected with at least one of each color?

Solution:

We count cases with at least one red, one green, one blue.

Approach: Use generating functions or inclusion-exclusion, but simpler to manually count valid (r, g, b) combinations that sum to 5 with each \(\geq 1\):

Possible (r, g, b) combinations:

- (3,1,1): \(\binom{6}{3} \cdot \binom{5}{1} \cdot \binom{4}{1} = 20 \cdot 5 \cdot 4 = 400\) - (2,2,1): \(\binom{6}{2} \cdot \binom{5}{2} \cdot \binom{4}{1} = 15 \cdot 10 \cdot 4 = 600\) - (2,1,2): \(15 \cdot 5 \cdot 6 = 450\) - (1,3,1): \(6 \cdot 10 \cdot 4 = 240\) - (1,2,2): \(6 \cdot 10 \cdot 6 = 360\) - (1,1,3): \(6 \cdot 5 \cdot 4 = 120\)

Total = \(400 + 600 + 450 + 240 + 360 + 120 = \boxed{2170}\)