Permutations and Combinations — NCERT Solutions

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

${}^nC_2 = 45$.

(i) ${}^{2n}C_3 : {}^nC_3 = 12 : 1$:

$$\frac{{}^{2n}C_3}{{}^nC_3} = 12$$

$$\frac{\dfrac{(2n)!}{3!(2n-3)!}}{\dfrac{n!}{3!(n-3)!}} = 12$$

$$\frac{(2n)(2n-1)(2n-2)}{n(n-1)(n-2)} = 12$$

$$\frac{2n \cdot (2n-1) \cdot 2(n-1)}{n(n-1)(n-2)} = 12$$

$$\frac{4(2n-1)}{n-2} = 12$$

$$4(2n-1) = 12(n-2)$$

$$8n - 4 = 12n - 24$$

$$20 = 4n$$

$$\boxed{n = 5}$$

(ii) ${}^{2n}C_3 : {}^nC_3 = 11 : 1$:

$$\frac{(2n)(2n-1)(2n-2)}{n(n-1)(n-2)} = 11$$

$$\frac{4(2n-1)}{n-2} = 11$$

$$4(2n-1) = 11(n-2)$$

$$8n - 4 = 11n - 22$$

$$18 = 3n$$

$$\boxed{n = 6}$$

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.