Permutations and Combinations — NCERT Solutions

Given: PERMUTATIONS (12 letters); starts with P, ends with S.

Letters: P, E, R, M, U, T, A, T, I, O, N, S — T appears twice.

Working:
Fix P at start and S at end. Arrange remaining 10 letters (with T repeated twice) in 10 middle positions:
$$\frac{10!}{2!} = \frac{3628800}{2} = 1814400$$

Answer: $1814400$ arrangements

Given: PERMUTATIONS; exactly 5 letters between P and S.

Working:
P and S with 5 letters between them can occupy positions:
$(1,7), (2,8), (3,9), (4,10), (5,11), (6,12)$ → 6 positions for the pair.
P and S can be arranged in 2 ways (PS or SP).

Choose 5 letters from remaining 10 (with T twice) for the middle positions and arrange them:
$$\frac{10!}{2!} \div \binom{10}{5}\text{ — use direct method:}$$

Arrange the 5 letters between P and S: choose 5 from remaining 10 letters (T,T,E,R,M,U,A,I,O,N) and arrange:

Total arrangements of all 12 letters with P and S having exactly 5 letters between them:
$$6 \times 2 \times \frac{10!}{2!} = 12 \times \frac{3628800}{2} = 12 \times 1814400 = 21772800$$

Answer: $21{,}772{,}800$ arrangements

Given: PERMUTATIONS; all vowels together.

Vowels: E, U, A, I, O → 5 vowels
Consonants: P, R, M, T, T, N, S → 7 consonants (T repeated twice)

Working:
Treat 5 vowels as one block → 8 units total (7 consonants + 1 block).

• Arrange 8 units (with T twice): $\dfrac{8!}{2!}$ ways
• Arrange 5 vowels within block: $5!$ ways

$$\text{Total} = \frac{8!}{2!} \times 5! = \frac{40320}{2} \times 120 = 20160 \times 120 = 2419200$$

Answer: $2419200$ arrangements

Given: LAUGHTER has 8 distinct letters; starts with L, does not end with R.

Working:
Fix L at start. Remaining 7 letters to fill positions 2–8.

• Total arrangements starting with L: $7! = 5040$
• Arrangements starting with L AND ending with R: $6! = 720$

$$\text{Required} = 7! - 6! = 5040 - 720 = 4320$$

Answer: $4320$ words

Given: LAUGHTER; no two vowels adjacent.

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

Working:
Arrange 5 consonants: $5!$ ways.
This creates 6 gaps: _ C _ C _ C _ C _ C _

Place 3 vowels in 6 gaps:
$${}^6P_3 = 6 \times 5 \times 4 = 120$$

$$\text{Total} = 5! \times {}^6P_3 = 120 \times 120 = 14400$$

Answer: $14400$ words

Given: LAUGHTER; relative positions of vowels and consonants unchanged.

Vowels: A, U, E (3 vowels at fixed relative positions)
Consonants: L, G, H, T, R (5 consonants at fixed relative positions)

Working:

• Arrange 3 vowels among themselves: $3! = 6$ ways
• Arrange 5 consonants among themselves: $5! = 120$ ways

$$\text{Total} = 3! \times 5! = 6 \times 120 = 720$$

Answer: $720$ words

Given: 5 Maths + 4 English + 3 Accountancy = 12 books; same subject books together.

Working:
Treat each subject as one block → 3 blocks.

• Arrange 3 blocks: $3! = 6$ ways
• Arrange 5 Maths books: $5!$ ways
• Arrange 4 English books: $4!$ ways
• Arrange 3 Accountancy books: $3!$ ways

$$\text{Total} = 3! \times 5! \times 4! \times 3! = 6 \times 120 \times 24 \times 6 = 103680$$

Answer: $103680$ ways

Given: 5 Maths + 4 English + 3 Accountancy = 12 books; no two books of same subject adjacent.

Working (using the given answer):
Arrange 5 Maths books in ${}^5P_5 = 5!$ ways.
Place 4 English books in the 6 gaps created: ${}^6P_4$ ways.
Place 3 Accountancy books in remaining gaps: ${}^{10}P_3$ ways (after placing 9 books, 10 gaps exist).

$$\text{Total} = {}^5P_5 \times {}^6P_4 \times {}^{10}P_3$$
$$= 120 \times 360 \times 720 = 31104000$$

Answer: $31{,}104{,}000$ ways

Given: Letters of LATE: A, E, L, T (alphabetical order).

Working:
Alphabetical order: A, E, L, T

• Words starting with A: $3! = 6$
• Words starting with E: $3! = 6$
• Words starting with LA: remaining E, T → $2! = 2$ words (LAET, LATE)
• LAET comes before LATE
• So LATE is at position: $6 + 6 + 1 + 1 = 14$

Answer: Rank of LATE = $14$

Given: AGAIN → letters: A, A, G, I, N (A repeated twice).

Total words:
$$\frac{5!}{2!} = \frac{120}{2} = 60$$

Dictionary arrangement (alphabetical order of letters: A, A, G, I, N):

Words starting with A: fix A, arrange A,G,I,N → $\dfrac{4!}{1!} = 24$ (A appears once in remaining)

Actually remaining letters after first A: A, G, I, N (A once) → $4! = 24$ words

Words starting with G: remaining A,A,I,N → $\dfrac{4!}{2!} = 12$ words

Words starting with I: remaining A,A,G,N → $\dfrac{4!}{2!} = 12$ words

Words starting with N: remaining A,A,G,I → $\dfrac{4!}{2!} = 12$ words

Cumulative: A(24), G(12→36), I(12→48), N(12→60)

49th word: First word starting with N:
N + arrange A,A,G,I alphabetically → NAAGI

50th word: Second word starting with N:
N + A + arrange A,G,I → NAAIG

Answer: Total = $60$ words; 49th word = NAAGI; 50th word = NAAIG

Given: From $(1,2)$ to $(7,5)$; steps: R (right) or U (up).

Working:

• Steps to the right: $7 - 1 = 6$ steps (R)
• Steps upward: $5 - 2 = 3$ steps (U)
• Total steps: $6 + 3 = 9$

Number of paths = number of arrangements of 6 R's and 3 U's:
$$\frac{9!}{6! \times 3!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = \frac{504}{6} = 84$$

Answer: $84$ paths

Given: Digits: 2, 3, 3, 5, 5, 6, 8 (7 digits total; 3 repeated twice, 5 repeated twice).

Working:
All 7-digit numbers using these digits:
$$\frac{7!}{2! \times 2!} = \frac{5040}{4} = 1260$$

Numbers greater than 5,000,000 must start with 5, 6, or 8.

Starting with 5: Remaining digits: 2,3,3,5,6,8 → $\dfrac{6!}{2!} = 360$

Starting with 6: Remaining digits: 2,3,3,5,5,8 → $\dfrac{6!}{2! \times 2!} = 180$

Starting with 8: Remaining digits: 2,3,3,5,5,6 → $\dfrac{6!}{2! \times 2!} = 180$

$$\text{Total} = 360 + 180 + 180 = 720$$

Answer: $720$ positive integers

Given: 10 members; 4 distinct posts to be filled.

Working:
$${}^{10}P_4 = \frac{10!}{6!} = 10 \times 9 \times 8 \times 7 = 5040$$

Answer: $5040$ ways

Given: 9 people; 2 specific people must sit adjacent; circular arrangement.

Working:
Treat the 2 insisting people as one unit → 8 units.

• Circular arrangements of 8 units: $(8-1)! = 7!$ ways
• The 2 people can swap within the unit: $2!$ ways

$$\text{Total} = 7! \times 2 = 5040 \times 2 = 10080$$

Answer: $10080$ ways

Given: ADINI → letters: A, D, I, I, N (I repeated twice).

Alphabetical order: A, D, I, I, N

Counting words:

• Starting with A: arrange D,I,I,N → $\dfrac{4!}{2!} = 12$ words (1–12)
• Starting with D: arrange A,I,I,N → $\dfrac{4!}{2!} = 12$ words (13–24)
• Starting with I: arrange A,D,I,N → $\dfrac{4!}{1!} = 24$ words (25–48)

So the 46th word is among words starting with I.

Words starting with I (positions 25–48): arrange A,D,I,N in alphabetical order.

• IA__: arrange D,I,N → $3! = 6$ words (25–30)
• ID__: arrange A,I,N → $3! = 6$ words (31–36)
• II__: arrange A,D,N → $3! = 6$ words (37–42)
• IN__: arrange A,D,I → $3! = 6$ words (43–48)

43rd: INADII → INADI
44th: INAID
45th: INDAI
46th: INDIA

Let me list words starting with IN (positions 43–48):
Remaining letters after IN: A, D, I
Alphabetical: A, D, I
43: I-N-A-D-I
44: I-N-A-I-D
45: I-N-D-A-I
46: I-N-D-I-A = INDIA

Answer: The 46th word is INDIA

Given: SCHOOL → letters: C, H, L, O, O, S (O repeated twice).

Alphabetical order: C, H, L, O, O, S

Counting words before SCHOOL:

• Starting with C: arrange H,L,O,O,S → $\dfrac{5!}{2!} = 60$
• Starting with H: arrange C,L,O,O,S → $\dfrac{5!}{2!} = 60$
• Starting with L: arrange C,H,O,O,S → $\dfrac{5!}{2!} = 60$
• Starting with O: arrange C,H,L,O,S → $5! = 120$

Subtotal before S: $60+60+60+120 = 300$

Now words starting with S:

• SC____: arrange H,L,O,O → $\dfrac{4!}{2!} = 12$ (but we need to check if SCHOOL starts with SC)

SCHOOL: S-C-H-O-O-L

• Starting with SC: arrange H,L,O,O → $\dfrac{4!}{2!} = 12$

But SCHOOL = S,C,H,O,O,L — check words before SCH:

• SCH___: remaining O,O,L

Words starting with SCHO: remaining O,L

• SCHOOL: S,C,H,O,O,L

Words starting with SCH: arrange O,O,L → $\dfrac{3!}{2!} = 3$

• SCHOOL (S,C,H,O,O,L) — this IS SCHOOL

Before SCHOOL within SCH: words starting with SCHL: arrange O,O → 1 word (SCHLOOL? No)

Let me redo carefully:

After S, next letter in SCHOOL is C. Letters remaining after S: C,H,L,O,O

• SA...: no A available, skip
• SC...: fix SC, remaining: H,L,O,O
• SCH...: fix SCH, remaining: L,O,O
• SCHL...: fix SCHL, remaining: O,O → 1 word: SCHLOO — wait, SCHOOL has O before L
• SCHO...: fix SCHO, remaining: L,O
• SCHOOL: S,C,H,O,O,L — fix SCHOO, remaining: L → 1 word: SCHOOL

Before SCHOOL within SCHO_: words starting with SCHL come before SCHO (L < O alphabetically)

• SCHL__: arrange O,O → $\dfrac{2!}{2!}=1$ word: SCHLOO — but that's 6 letters: S,C,H,L,O,O = SCHLОО

So 1 word before SCHOOL within SCH group.
Words starting with SCH before SCHOOL: 1 (SCHLOO)
Words starting with SC before SCH: none (H is next letter after C in SCHOOL, and H comes after C alphabetically — no letters between C and H from {H,L,O,O})

Actually within SC_, letters available are H,L,O,O. Alphabetical: H,L,O,O.

• SCH: this is our path

Before SCH: no letters come before H from {H,L,O,O} — H is first.

Within SCH_, remaining: L,O,O. Alphabetical: L,O,O

• SCHL: before SCHO (L < O)

SCHL + OO: 1 arrangement: SCHLOO → 1 word

• SCHO: our path

Within SCHO_, remaining: L,O

• SCHOL: before SCHOO (L < O)

SCHOL + O: SCHOLО → 1 word

• SCHOO: our path

SCHOOL: only 1 arrangement → this IS SCHOOL

Rank = 300 (before S) + 0 (SC before SCH) + 1 (SCHLOO) + 1 (SCHOLO) + 1 (SCHOOL itself)
= 300 + 1 + 1 + 1 = 303

Answer: Rank of SCHOOL = $\mathbf{303}$

Given: ${}^{18}C_r = {}^{18}C_{r+2}$

Using property: ${}^nC_x = {}^nC_y \Rightarrow x + y = n$ (when $x \neq y$)

$$r + (r+2) = 18 \Rightarrow 2r + 2 = 18 \Rightarrow r = 8$$

$${}^rC_5 = {}^8C_5 = {}^8C_3 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$$

Answer: ${}^rC_5 = 56$

Given: $\dfrac{{}^nC_r}{{}^nC_{r+1}} = \dfrac{1}{2}$ and $\dfrac{{}^nC_{r+1}}{{}^nC_{r+2}} = \dfrac{2}{3}$

Using: $\dfrac{{}^nC_r}{{}^nC_{r+1}} = \dfrac{r+1}{n-r}$

From first ratio:
$$\frac{r+1}{n-r} = \frac{1}{2} \Rightarrow 2(r+1) = n-r \Rightarrow 2r+2 = n-r \Rightarrow n = 3r+2 \quad \cdots(1)$$

From second ratio:
$$\frac{r+2}{n-r-1} = \frac{2}{3} \Rightarrow 3(r+2) = 2(n-r-1)$$
$$3r+6 = 2n-2r-2 \Rightarrow 5r+8 = 2n \quad \cdots(2)$$

Substituting (1) into (2):
$$5r+8 = 2(3r+2) = 6r+4$$
$$8-4 = 6r-5r \Rightarrow r = 4$$
$$n = 3(4)+2 = 14$$

Answer: $n = 14,\ r = 4$

Hint given: Write ${}^nC_0 = {}^{n+1}C_0$

Proof by using Pascal's identity: ${}^mC_k + {}^mC_{k-1} = {}^{m+1}C_k$

Working:
$${}^nC_0 + {}^{n+1}C_1 + {}^{n+2}C_2 + \cdots + {}^{n+r}C_r$$

Since ${}^nC_0 = {}^{n+1}C_0$:
$$= {}^{n+1}C_0 + {}^{n+1}C_1 + {}^{n+2}C_2 + \cdots + {}^{n+r}C_r$$

Apply Pascal's identity: ${}^{n+1}C_0 + {}^{n+1}C_1 = {}^{n+2}C_1$:
$$= {}^{n+2}C_1 + {}^{n+2}C_2 + {}^{n+3}C_3 + \cdots + {}^{n+r}C_r$$

Continuing:
$$= {}^{n+3}C_2 + {}^{n+3}C_3 + \cdots + {}^{n+r}C_r$$

Repeating this process $r$ times:
$$= {}^{n+r+1}C_r \quad \blacksquare$$

Given: 17 points on a circle; a chord is determined by any 2 points.

Working:
$$\text{Number of chords} = {}^{17}C_2 = \frac{17 \times 16}{2} = 136$$

Answer: $136$ chords

Given: Diagonals = $2 \times$ sides.

Formula: Number of diagonals of an $n$-sided polygon $= {}^nC_2 - n = \dfrac{n(n-1)}{2} - n = \dfrac{n(n-3)}{2}$

Setting up equation:
$$\frac{n(n-3)}{2} = 2n$$
$$n(n-3) = 4n$$
$$n-3 = 4 \quad (n \neq 0)$$
$$n = 7$$

Answer: The polygon has $7$ sides (heptagon).

Given: 6 red, 7 white balls; select 4 red and 3 white.

Working:
$${}^6C_4 \times {}^7C_3 = \frac{6!}{4! \cdot 2!} \times \frac{7!}{3! \cdot 4!} = 15 \times 35 = 525$$

Answer: $525$ ways

Given: 4 teachers, 6 students; committee of 5 with at least 2 students.

Cases:

$$\text{Total} = 60 + 120 + 60 + 6 = 246$$

Answer: $246$ ways

Given: 15 players; 2 specific players must be included; select 11.

Working:
With 2 fixed, choose remaining 9 from 13:
$${}^{13}C_9 = {}^{13}C_4 = \frac{13 \times 12 \times 11 \times 10}{4 \times 3 \times 2 \times 1} = \frac{17160}{24} = 715$$

Answer: $715$ ways

Given: 15 players; 1 specific player excluded; select 11 from remaining 14.

Working:
$${}^{14}C_{11} = {}^{14}C_3 = \frac{14 \times 13 \times 12}{3 \times 2 \times 1} = \frac{2184}{6} = 364$$

Answer: $364$ ways

Given: 10 courses; 2 language courses compulsory; choose 5 total.

Working:
2 courses are fixed. Choose remaining 3 from remaining 8 courses:
$${}^8C_3 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$$

Answer: $56$ ways

Given: 7 (+) signs and 5 (–) signs; no two (–) signs adjacent.

Working:
First arrange 7 (+) signs: creates 8 gaps (including ends).
_ + _ + _ + _ + _ + _ + _ + _

Place 5 (–) signs in 8 gaps (at most one per gap):
$${}^8C_5 = {}^8C_3 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$$

Answer: $56$ ways

Given: 20 points in space; no four coplanar.

Triangles: Any 3 points determine a unique triangle.
$${}^{20}C_3 = \frac{20 \times 19 \times 18}{3 \times 2 \times 1} = 1140$$

Planes: Any 3 non-collinear points determine a unique plane. Since no four are coplanar, any 3 points give a distinct plane.
$${}^{20}C_3 = 1140$$

Tetrahedrons: Any 4 points (no four coplanar) determine a unique tetrahedron.
$${}^{20}C_4 = \frac{20 \times 19 \times 18 \times 17}{4 \times 3 \times 2 \times 1} = \frac{116280}{24} = 4845$$