Permutations and Combinations

Given: Digits available: 1, 2, 3, 4, 5 (total 5 digits). We need to form 3-digit numbers.

(i) Repetition allowed:

Each of the 3 places (hundreds, tens, units) can be filled by any of the 5 digits.

By the Fundamental Principle of Counting:
$$\text{Number of 3-digit numbers} = 5 \times 5 \times 5 = 125$$

(ii) Repetition not allowed:

- Hundreds place: 5 choices
- Tens place: 4 choices (one digit used)
- Units place: 3 choices (two digits used)

$$\text{Number of 3-digit numbers} = 5 \times 4 \times 3 = 60$$

Given: Digits: 1, 2, 3, 4, 5, 6. Repetition is allowed. The number must be even.

For a number to be even, the units digit must be even: 2, 4, or 6 → 3 choices.

- Units place: 3 choices (2, 4, or 6)
- Tens place: 6 choices (any digit, repetition allowed)
- Hundreds place: 6 choices

By the Fundamental Principle of Counting:
$$\text{Number of 3-digit even numbers} = 6 \times 6 \times 3 = 108$$

Given: First 10 letters of the English alphabet (A, B, C, D, E, F, G, H, I, J). No repetition allowed. Code has 4 letters.

This is the number of permutations of 10 letters taken 4 at a time:
$${}^{10}P_4 = 10 \times 9 \times 8 \times 7 = 5040$$

Therefore, 5040 four-letter codes can be formed.

Given: 5-digit telephone numbers starting with 67. Digits 0–9 (10 digits). No repetition.

Since the number starts with 67, the first two digits are fixed: 6 and 7.

Remaining digits available: 10 − 2 = 8 digits (0,1,2,3,4,5,8,9).

We need to fill 3 more places (3rd, 4th, 5th digits) from these 8 digits without repetition:

- 3rd place: 8 choices
- 4th place: 7 choices
- 5th place: 6 choices

$$\text{Number of telephone numbers} = 8 \times 7 \times 6 = 336$$

Given: A coin is tossed 3 times. Each toss has 2 possible outcomes: Head (H) or Tail (T).

By the Fundamental Principle of Counting:
$$\text{Total outcomes} = 2 \times 2 \times 2 = 8$$

There are 8 possible outcomes.

Given: 5 flags of different colours. Each signal uses exactly 2 flags, one below the other (order matters).

This is the number of permutations of 5 flags taken 2 at a time:
$${}^5P_2 = 5 \times 4 = 20$$

Therefore, 20 different signals can be generated.

(i) $8! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 = 40320$

(ii) $4! = 1 \times 2 \times 3 \times 4 = 24$

$3! = 1 \times 2 \times 3 = 6$

$$4! - 3! = 24 - 6 = 18$$

Calculating each side:

$3! = 6$

$4! = 24$

$3! + 4! = 6 + 24 = 30$

$7! = 5040$

Since $30 \neq 5040$,

$$3! + 4! \neq 7!$$

No, $3! + 4! \neq 7!$.

Given: $\dfrac{1}{6!} + \dfrac{1}{7!} = \dfrac{x}{8!}$

Rewrite using $7! = 7 \times 6!$ and $8! = 8 \times 7 \times 6!$:

$$\frac{1}{6!} + \frac{1}{7 \times 6!} = \frac{x}{8 \times 7 \times 6!}$$

Divide throughout by $\dfrac{1}{6!}$:

$$1 + \frac{1}{7} = \frac{x}{56}$$

$$\frac{8}{7} = \frac{x}{56}$$

$$x = \frac{8 \times 56}{7} = 8 \times 8 = 64$$

Therefore, $x = 64$.

Formula: $\dfrac{n!}{(n-r)!} = n(n-1)(n-2)\cdots(n-r+1)$

(i) $n = 6, r = 2$:
$$\frac{6!}{(6-2)!} = \frac{6!}{4!} = 6 \times 5 = 30$$

(ii) $n = 9, r = 5$:
$$\frac{9!}{(9-5)!} = \frac{9!}{4!} = 9 \times 8 \times 7 \times 6 \times 5 = 15120$$

Given: Digits 1 to 9 (9 digits). 3-digit numbers, no repetition.

This is ${}^9P_3$:
$${}^9P_3 = 9 \times 8 \times 7 = 504$$

504 three-digit numbers can be formed.

Given: Digits 0–9 (10 digits). 4-digit numbers, no repetition.

The thousands place cannot be 0, so:
- Thousands place: 9 choices (1–9)
- Hundreds place: 9 choices (0 and remaining 8 digits)
- Tens place: 8 choices
- Units place: 7 choices

$$\text{Total} = 9 \times 9 \times 8 \times 7 = 4536$$

There are 4536 four-digit numbers with no digit repeated.

Given: Digits: 1, 2, 3, 4, 6, 7. No repetition. Number must be even.

Even digits available: 2, 4, 6 → units place has 3 choices.

After fixing units digit:
- Hundreds place: 5 choices (from remaining 5 digits)
- Tens place: 4 choices

$$\text{Total} = 5 \times 4 \times 3 = 60$$

60 three-digit even numbers can be formed.

Given: Digits: 1, 2, 3, 4, 5. No repetition.

Total 4-digit numbers:
$${}^5P_4 = 5 \times 4 \times 3 \times 2 = 120$$

Even 4-digit numbers:

For the number to be even, units digit must be 2 or 4 → 2 choices.

Remaining 3 places filled from remaining 4 digits:
$${}^4P_3 = 4 \times 3 \times 2 = 24$$

$$\text{Even numbers} = 2 \times 24 = 48$$

Total 4-digit numbers = 120; Even 4-digit numbers = 48.

Given: 8 persons. Choose 1 chairman and 1 vice chairman (different persons, order matters).

- Chairman: 8 choices
- Vice Chairman: 7 choices (remaining persons)

$$\text{Number of ways} = 8 \times 7 = 56$$

The chairman and vice chairman can be chosen in 56 ways.

Given: ${}^{n-1}P_3 : {}^nP_4 = 1 : 9$

$$\frac{{}^{n-1}P_3}{{}^nP_4} = \frac{1}{9}$$

$$\frac{(n-1)!}{(n-1-3)!} \div \frac{n!}{(n-4)!} = \frac{1}{9}$$

$$\frac{(n-1)!}{(n-4)!} \times \frac{(n-4)!}{n!} = \frac{1}{9}$$

$$\frac{(n-1)!}{n!} = \frac{1}{9}$$

$$\frac{1}{n} = \frac{1}{9}$$

$$\boxed{n = 9}$$

(i) ${}^5P_r = 2 \cdot {}^6P_{r-1}$:

$$\frac{5!}{(5-r)!} = 2 \cdot \frac{6!}{(6-(r-1))!} = 2 \cdot \frac{6!}{(7-r)!}$$

$$\frac{5!}{(5-r)!} = \frac{2 \times 6!}{(7-r)!}$$

$$\frac{5!}{(5-r)!} = \frac{2 \times 6 \times 5!}{(7-r)(6-r)(5-r)!}$$

Divide both sides by $\dfrac{5!}{(5-r)!}$:

$$1 = \frac{12}{(7-r)(6-r)}$$

$$(7-r)(6-r) = 12$$

$$42 - 13r + r^2 = 12$$

$$r^2 - 13r + 30 = 0$$

$$(r-3)(r-10) = 0$$

$r = 3$ or $r = 10$. Since $r \leq 5$ (for ${}^5P_r$ to be defined), $r = 10$ is rejected.

$$\boxed{r = 3}$$

(ii) ${}^5P_r = {}^6P_{r-1}$:

$$\frac{5!}{(5-r)!} = \frac{6!}{(7-r)!}$$

$$\frac{5!}{(5-r)!} = \frac{6 \times 5!}{(7-r)(6-r)(5-r)!}$$

Divide both sides by $\dfrac{5!}{(5-r)!}$:

$$1 = \frac{6}{(7-r)(6-r)}$$

$$(7-r)(6-r) = 6$$

$$42 - 13r + r^2 = 6$$

$$r^2 - 13r + 36 = 0$$

$$(r-4)(r-9) = 0$$

$r = 4$ or $r = 9$. Since $r \leq 5$, $r = 9$ is rejected.

$$\boxed{r = 4}$$

Given: Word EQUATION has 8 letters: E, Q, U, A, T, I, O, N — all distinct.

Number of arrangements of 8 distinct letters:
$$8! = 40320$$

40320 words can be formed.

Given: MONDAY has 6 distinct letters: M, O, N, D, A, Y. Vowels: O, A (2 vowels); Consonants: M, N, D, Y (4 consonants).

(i) 4 letters at a time:
$${}^6P_4 = 6 \times 5 \times 4 \times 3 = 360$$

(ii) All 6 letters at a time:
$$6! = 720$$

(iii) All 6 letters, first letter is a vowel:

- First place: 2 choices (O or A)
- Remaining 5 places: $5!$ arrangements of remaining 5 letters

$$\text{Total} = 2 \times 5! = 2 \times 120 = 240$$

Given: MISSISSIPPI has 11 letters: M(1), I(4), S(4), P(2).

Total distinct permutations:
$$\frac{11!}{1! \cdot 4! \cdot 4! \cdot 2!} = \frac{39916800}{24 \times 24 \times 2} = \frac{39916800}{1152} = 34650$$

Permutations where all four I's come together:

Treat IIII as one unit → 8 units: M, (IIII), S, S, S, S, P, P

$$\frac{8!}{4! \cdot 2!} = \frac{40320}{24 \times 2} = \frac{40320}{48} = 840$$

Permutations where four I's do NOT come together:
$$34650 - 840 = 33810$$

Given: PERMUTATIONS has 12 letters: P, E, R, M, U, T, A, T, I, O, N, S. The letter T is repeated twice; all others appear once.

Total letters = 12, with T repeated 2 times.

(i) Words start with P and end with S:

Fix P at the first position and S at the last. Remaining 10 letters (including T twice) arranged in 10 places:
$$\frac{10!}{2!} = \frac{3628800}{2} = 1814400$$

(ii) Vowels are all together:

Vowels in PERMUTATIONS: E, U, A, I, O → 5 vowels (all distinct).

Treat the 5 vowels as one unit. Now we have 8 units: (EUAIO), P, R, M, T, T, N, S.

Arrangements of these 8 units (T repeated twice):
$$\frac{8!}{2!} = \frac{40320}{2} = 20160$$

The 5 vowels within the unit can be arranged in:
$$5! = 120 \text{ ways}$$

$$\text{Total} = 20160 \times 120 = 2419200$$

(iii) There are always 4 letters between P and S:

We need to find positions for P and S such that exactly 4 letters lie between them.

Possible position pairs (P at position $i$, S at position $j$, with $|i - j| = 5$):

Pairs where difference is 5: (1,6),(2,7),(3,8),(4,9),(5,10),(6,11),(7,12) → 7 pairs.

P and S can be interchanged, so total arrangements of P and S = $7 \times 2 = 14$.

Remaining 10 letters (with T twice) fill the other 10 positions:
$$\frac{10!}{2!} = 1814400$$

$$\text{Total} = 14 \times 1814400 = 25401600$$

Given: ${}^nC_8 = {}^nC_2$

Using the property ${}^nC_r = {}^nC_{n-r}$:

If ${}^nC_8 = {}^nC_2$, then either $8 = 2$ (impossible) or $8 = n - 2$, giving $n = 10$.

Now:
$${}^{10}C_2 = \frac{10!}{2! \cdot 8!} = \frac{10 \times 9}{2} = 45$$

${}^nC_2 = 45$.

Given: 21 points on a circle. A chord is determined by choosing any 2 points.

$$\text{Number of chords} = {}^{21}C_2 = \frac{21 \times 20}{2} = 210$$

210 chords can be drawn.

Given: 5 boys, 4 girls. Select 3 boys and 3 girls.

$$\text{Ways to select 3 boys from 5} = {}^5C_3 = \frac{5!}{3! \cdot 2!} = 10$$

$$\text{Ways to select 3 girls from 4} = {}^4C_3 = \frac{4!}{3! \cdot 1!} = 4$$

$$\text{Total ways} = 10 \times 4 = 40$$

The team can be selected in 40 ways.

Given: 6 red, 5 white, 5 blue balls. Select 3 of each colour.

$$\text{Red: } {}^6C_3 = \frac{6!}{3! \cdot 3!} = 20$$

$$\text{White: } {}^5C_3 = \frac{5!}{3! \cdot 2!} = 10$$

$$\text{Blue: } {}^5C_3 = 10$$

$$\text{Total} = 20 \times 10 \times 10 = 2000$$

The number of ways is 2000.

Given: 52 cards (4 aces, 48 non-aces). Select 5 cards with exactly 1 ace.

- Choose 1 ace from 4: ${}^4C_1 = 4$
- Choose 4 non-ace cards from 48: ${}^{48}C_4 = \dfrac{48 \times 47 \times 46 \times 45}{4!} = \dfrac{4669920}{24} = 194580$

$$\text{Total} = 4 \times 194580 = 778320$$

The number of 5-card combinations is 778320.

Given: 17 players: 5 bowlers, 12 non-bowlers. Team of 11 with exactly 4 bowlers.

- Choose 4 bowlers from 5: ${}^5C_4 = 5$
- Choose 7 non-bowlers from 12: ${}^{12}C_7 = \dfrac{12!}{7! \cdot 5!} = 792$

$$\text{Total} = 5 \times 792 = 3960$$

The cricket team can be selected in 3960 ways.

Given: 5 black balls, 6 red balls. Select 2 black and 3 red.

$${}^5C_2 = \frac{5!}{2! \cdot 3!} = 10$$

$${}^6C_3 = \frac{6!}{3! \cdot 3!} = 20$$

$$\text{Total} = 10 \times 20 = 200$$

The balls can be selected in 200 ways.

Given: 9 courses total, 2 are compulsory. Student must choose 5 courses.

Since 2 courses are compulsory, the student needs to choose $5 - 2 = 3$ more courses from the remaining $9 - 2 = 7$ courses.

$${}^7C_3 = \frac{7!}{3! \cdot 4!} = \frac{7 \times 6 \times 5}{6} = 35$$

The student can choose the programme in 35 ways.

Given: DAUGHTER: D, A, U, G, H, T, E, R

Vowels: A, U, E → 3 vowels
Consonants: D, G, H, T, R → 5 consonants

Step 1: Choose 2 vowels from 3:
$${}^3C_2 = 3$$

Step 2: Choose 3 consonants from 5:
$${}^5C_3 = 10$$

Step 3: Arrange the selected 5 letters:
$$5! = 120$$

Total words:
$$3 \times 10 \times 120 = 3600$$

3600 words can be formed.

Given: EQUATION: E, Q, U, A, T, I, O, N (8 letters)

Vowels: E, U, A, I, O → 5 vowels
Consonants: Q, T, N → 3 consonants

For vowels and consonants to occur together, treat all vowels as one group and all consonants as one group.

Step 1: Arrange the 2 groups:
$$2! = 2$$

Step 2: Arrange 5 vowels within their group:
$$5! = 120$$

Step 3: Arrange 3 consonants within their group:
$$3! = 6$$

Total:
$$2 \times 120 \times 6 = 1440$$

1440 words can be formed.

Given: 9 boys, 4 girls. Committee of 7.

(i) Exactly 3 girls:

Choose 3 girls from 4 and 4 boys from 9:
$${}^4C_3 \times {}^9C_4 = 4 \times 126 = 504$$

(ii) At least 3 girls (3 or 4 girls):

- Exactly 3 girls: ${}^4C_3 \times {}^9C_4 = 4 \times 126 = 504$
- Exactly 4 girls: ${}^4C_4 \times {}^9C_3 = 1 \times 84 = 84$

$$\text{Total} = 504 + 84 = 588$$

(iii) At most 3 girls (0, 1, 2, or 3 girls):

Total ways to form committee of 7 from 13 people:
$${}^{13}C_7 = 1716$$

At most 3 girls = Total − (at least 4 girls) = Total − (exactly 4 girls)

$$= 1716 - 84 = 1632$$

Answers: (i) 504, (ii) 588, (iii) 1632.

Given: EXAMINATION has 11 letters: E, X, A, M, I, N, A, T, I, O, N

Letter frequencies: A(2), I(2), N(2), E(1), X(1), M(1), T(1), O(1). Total = 11 letters.

Words before those starting with E are words starting with A.

Words starting with A:

Fix A at first position. Remaining 10 letters: E, X, A, M, I, N, T, I, O, N
Frequencies in remaining: I(2), N(2), and E, X, A, M, T, O each once.

$$\text{Arrangements} = \frac{10!}{2! \cdot 2!} = \frac{3628800}{4} = 907200$$

There are 907200 words before the first word starting with E.

Given: Digits: 0, 1, 3, 5, 7, 9. 6-digit numbers, no repetition, divisible by 10.

For divisibility by 10, the units digit must be 0.

Fix 0 at the units place. Remaining 5 digits (1, 3, 5, 7, 9) fill the first 5 places.

The first place (ten-thousands/leftmost) can be any of the 5 remaining digits (all non-zero, so no restriction).

$$\text{Arrangements of remaining 5 digits} = 5! = 120$$

120 six-digit numbers can be formed.

Given: 5 vowels, 21 consonants. Words with 2 different vowels and 2 different consonants.

Step 1: Choose 2 vowels from 5:
$${}^5C_2 = 10$$

Step 2: Choose 2 consonants from 21:
$${}^{21}C_2 = \frac{21 \times 20}{2} = 210$$

Step 3: Arrange the 4 selected letters:
$$4! = 24$$

Total:
$$10 \times 210 \times 24 = 50400$$

50400 words can be formed.

Given: Part I: 5 questions, Part II: 7 questions. Attempt 8 questions with at least 3 from each part.

Possible distributions (Part I : Part II):
- 3 from Part I and 5 from Part II
- 4 from Part I and 4 from Part II
- 5 from Part I and 3 from Part II

Case 1: 3 from Part I, 5 from Part II:
$${}^5C_3 \times {}^7C_5 = 10 \times 21 = 210$$

Case 2: 4 from Part I, 4 from Part II:
$${}^5C_4 \times {}^7C_4 = 5 \times 35 = 175$$

Case 3: 5 from Part I, 3 from Part II:
$${}^5C_5 \times {}^7C_3 = 1 \times 35 = 35$$

Total:
$$210 + 175 + 35 = 420$$

The student can select questions in 420 ways.

Given: 52 cards (4 kings, 48 non-kings). Select 5 cards with exactly 1 king.

- Choose 1 king from 4: ${}^4C_1 = 4$
- Choose 4 non-king cards from 48: ${}^{48}C_4 = \dfrac{48 \times 47 \times 46 \times 45}{24} = 194580$

$$\text{Total} = 4 \times 194580 = 778320$$

The number of 5-card combinations is 778320.

Given: 5 men and 4 women in a row of 9 seats. Women occupy even places.

Even places in a row of 9: positions 2, 4, 6, 8 → 4 even places.

- Arrange 4 women in 4 even places: $4! = 24$ ways
- Arrange 5 men in 5 odd places (1, 3, 5, 7, 9): $5! = 120$ ways

$$\text{Total arrangements} = 4! \times 5! = 24 \times 120 = 2880$$

2880 arrangements are possible.

Given: 25 students, choose 10. 3 specific students: either all join or none join.

Case 1: All 3 join.

Choose remaining 7 from the other 22 students:
$${}^{22}C_7 = 170544$$

Case 2: None of the 3 join.

Choose all 10 from the remaining 22 students:
$${}^{22}C_{10} = 646646$$

Total:
$$170544 + 646646 = 817190$$

The excursion party can be chosen in 817190 ways.

Given: ASSASSINATION has 13 letters.

Letter frequencies: A(3), S(4), I(2), N(2), T(1), O(1).

Treat all 4 S's as a single unit (SSSS). Now we have 10 units:
(SSSS), A, A, A, I, I, N, N, T, O

Frequencies in remaining: A(3), I(2), N(2), T(1), O(1).

$$\text{Number of arrangements} = \frac{10!}{3! \cdot 2! \cdot 2!} = \frac{3628800}{6 \times 2 \times 2} = \frac{3628800}{24} = 151200$$

The letters can be arranged in 151200 ways.