Exercise 6.1
How many 3-digit numbers can be formed from digits 1, 2, 3, 4, 5 if (i) repetition allowed (ii) not allowed?
Answer
(i) Each place can be filled in 5 ways: 5 × 5 × 5 = 125
(ii) Without repetition: 5 × 4 × 3 = 60
How many 3-digit even numbers can be formed from 1, 2, 3, 4, 5, 6 if digits can be repeated?
Answer
Units digit must be even: 2, 4, or 6
⇒ 3 choices. Tens and hundreds: 6 choices each.
Total = 6 × 6 × 3 = 108
How many 4-letter codes can be formed using first 10 letters of English alphabet if no letter repeated?
Answer
¹⁰P₄ = 10 × 9 × 8 × 7 = 5040
How many 5-digit telephone numbers starting with 67, using digits 0–9 with no digit repeated?
Answer
First two digits are fixed as 6 and 7. Remaining 3 places from remaining 8 digits: 8 × 7 × 6 = 336
A coin tossed 3 times. How many possible outcomes?
Answer
Each toss has 2 outcomes (H or T). Total = 2 × 2 × 2 = 8
5 flags of different colours. How many different signals using 2 flags (one below the other)?
Answer
⁵P₂ = 5 × 4 = 20
Exercise 6.2
Evaluate: (i) 8! (ii) 4! – 3!
Answer
(i) 8! = 40320
(ii) 4! – 3! = 24 – 6 = 18
Is 3! + 4! = 7!?
Answer
3! + 4! = 6 + 24 = 30. 7! = 5040. So 3! + 4! ≠ 7!. Answer: No.
Compute 8!/(6! × 2!).
Answer
8!/(6! × 2!) = (8 × 7 × 6!)/(6! × 2) = 56/2 = 28
If 1/6! + 1/7! = x/8!, find x.
Answer
1/6! + 1/7! = x/8!
8!/(6!) + 8!/(7!) = x
⇒ 8 × 7 + 8 = x
⇒ 56 + 8 = x
⇒ x = 64
Evaluate n!/(n–r)! when (i) n=6, r=2 (ii) n=9, r=5.
Answer
(i) 6!/4! = 6 × 5 = 30
(ii) 9!/4! = 9 × 8 × 7 × 6 × 5 = 15120
Exercise 6.3
How many 3-digit numbers from digits 1 to 9 with no digit repeated?
Answer
⁹P₃ = 9 × 8 × 7 = 504
How many 4-digit numbers with no digit repeated?
Answer
Thousands digit: 1–9 (9 choices). Remaining 3 from 9 digits (including 0): 9 × 9 × 8 × 7 = ...
More precisely: total 4-digit numbers = ⁹P₁ × ⁹P₃ for first digit non-zero approach.
= 9 × 9 × 8 × 7 = 4536
How many 3-digit even numbers from 1, 2, 3, 4, 6, 7 with no digit repeated?
Answer
Units digit: 2, 4, or 6 (3 choices). Remaining 2 places from 5 digits: 5 × 4 = 20 each.
Total = 3 × 5 × 4 = 60
4-digit numbers from 1, 2, 3, 4, 5 (no repetition). How many are even?
Answer
Total 4-digit: ⁵P₄ = 5! = 120
Even: units digit = 2 or 4 (2 choices). Remaining 3 places from 4 digits: ⁴P₃ = 24.
Even 4-digit numbers = 2 × 24 = 48
From committee of 8, in how many ways can chairman and vice-chairman be chosen (one person can't hold both)?
Answer
⁸P₂ = 8 × 7 = 56
Find n if ⁽ⁿ⁻¹⁾P₃ : ⁿP₄ = 1 : 9.
Answer
(n–1)!/(n–4)! ÷ n!/(n–4)! = 1/9
(n–1)!/n! = 1/9 → 1/n = 1/9 → n = 9
Find r if (i) ⁵Pᵣ = 2 × ⁶Pᵣ₋₁ (ii) ⁵Pᵣ = ⁶Pᵣ₋₁.
Answer
(i) 5!/(5–r)! = 2 × 6!/(7–r)!
⇒ 5!(7–r)! = 2 × 6!(5–r)!
5! × (7–r)(6–r) = 2 × 6 × 5!
⇒ (7–r)(6–r) = 12
⇒ r² – 13r + 42 – 12 = 0
⇒ r² – 13r + 30 = 0
→ (r–3)(r–10) = 0 → r = 3 (since r ≤ 5)
(ii) 5!/(5–r)! = 6!/(7–r)!
⇒ 5!(7–r)(6–r) = 6!
⇒ (7–r)(6–r) = 6
⇒ r=4
How many words (with or without meaning) using all letters of EQUATION (each once)?
Answer
EQUATION has 8 letters, all distinct. Number = 8! = 40320
Letters of MONDAY (no repetition): (i) 4 letters at a time (ii) all letters (iii) all letters but first is vowel.
Answer
(i) ⁶P₄ = 6 × 5 × 4 × 3 = 360
(ii) 6! = 720
(iii) Vowels in MONDAY: O, A (2 vowels). First letter from {O, A}: 2 choices. Remaining 5: 5! = 120. Total = 2 × 120 = 240
In distinct permutations of MISSISSIPPI, how many arrangements have the four I's not together?
Answer
MISSISSIPPI: M=1, I=4, S=4, P=2
⇒ total = 11!/(4!4!2!) = 34650
I's together: treat IIII as one unit
⇒ 8 objects: M, IIII, S, S, S, S, P, P
= 8!/(4!2!) = 840
I's NOT together = 34650 – 840 = 33810
Letters of PERMUTATIONS arranged: (i) starts with P, ends with S (ii) vowels together (iii) always 4 letters between P and S.
Answer
PERMUTATIONS: 12 letters, T repeated twice.
(i) P fixed first, S fixed last. Remaining 10 letters (with T twice): 10!/2! = 1814400
(ii) Vowels: E, U, A, I, O (5 vowels). Treat as unit. Unit + 7 consonants (T twice) = 8 items: 8!/2! × 5! = 20160 × 120 = 2419200
(iii) P and S with exactly 4 letters between: treat PS block (with 4 letters between) as taking 6 consecutive positions. Number of ways to place such a block in 12 positions = 7 positions (1–6, 2–7, ..., 7–12) × 2 (PS or SP).
= 7 × 2 × 10!/2! = 14 × 1814400 = 25401600
Exercise 6.4
If ⁿC₈ = ⁿC₂, find ⁿC₂.
Answer
ⁿC₈ = ⁿC₂ → n = 8+2 = 10
¹⁰C₂ = 10!/(2! × 8!) = 45
Determine n if (i) ²ⁿC₃ : ⁿC₃ = 12 : 1 (ii) ²ⁿC₃ : ⁿC₃ = 11 : 1.
Answer
(i) (2n)!/(3!(2n–3)!) ÷ n!/(3!(n–3)!) = 12
(2n)(2n–1)(2n–2)/[n(n–1)(n–2)] = 12
⇒ 2(2n–1) × 2(n–1)/[(n–1)(n–2)] = 12
4(2n–1)/(n–2) = 12
⇒ 2n–1 = 3(n–2)
⇒ 2n–1 = 3n–6
⇒ n = 5
(ii) (2n)(2n–1)(2n–2)/[n(n–1)(n–2)] = 11
⇒ 4(2n–1)/(n–2) = 11
⇒ 8n–4 = 11n–22
⇒ 3n = 18
⇒ n = 6
How many chords can be drawn through 21 points on a circle?
Answer
Each chord needs 2 points: ²¹C₂ = 21×20/2 = 210
A team of 3 boys and 3 girls from 5 boys and 4 girls.
Answer
⁵C₃ × ⁴C₃ = 10 × 4 = 40
Select 9 balls from 6 red, 5 white, 5 blue (3 of each colour).
Answer
⁶C₃ × ⁵C₃ × ⁵C₃ = 20 × 10 × 10 = 2000
5-card combinations from 52 cards with exactly one ace.
Answer
Choose 1 ace from 4: ⁴C₁ = 4. Choose 4 from 48 non-aces: ⁴⁸C₄ = 194580.
Total = 4 × 194580 = 778320
Cricket team of 11 from 17 players (5 bowlers), each team must include exactly 4 bowlers.
Answer
Choose 4 bowlers from 5: ⁵C₄ = 5. Choose 7 from 12 non-bowlers: ¹²C₇ = 792.
Total = 5 × 792 = 3960
Bag has 5 black and 6 red balls. Select 2 black and 3 red.
Answer
⁵C₂ × ⁶C₃ = 10 × 20 = 200
Student chooses 5 courses from 9, with 2 specific compulsory.
Answer
2 specific courses are fixed. Choose remaining 3 from 7: ⁷C₃ = 35
Miscellaneous Exercise
Words with 2 vowels and 3 consonants from DAUGHTER.
Answer
DAUGHTER: vowels A, U, E (3), consonants D, G, H, T, R (5).
Select 2 vowels from 3: ³C₂ = 3. Select 3 consonants from 5: ⁵C₃ = 10.
Arrange 5 letters: 5! = 120.
Total = 3 × 10 × 120 = 3600
Words using all letters of EQUATION with vowels and consonants together.
Answer
EQUATION: vowels E, U, A, I, O (5), consonants Q, T, N (3).
Treat vowel group and consonant group as 2 units: 2! ways. Vowels among themselves: 5!. Consonants: 3!.
Total = 2! × 5! × 3! = 2 × 120 × 6 = 1440
Committee of 7 from 9 boys and 4 girls: (i) exactly 3 girls (ii) at least 3 girls (iii) at most 3 girls.
Answer
(i) ⁴C₃ × ⁹C₄ = 4 × 126 = 504
(ii) At least 3 girls: exactly 3 or exactly 4 girls.
= ⁴C₃×⁹C₄ + ⁴C₄×⁹C₃ = 504 + 1×84 = 588
(iii) At most 3 girls: 0, 1, 2, or 3 girls. Total ways = ¹³C₇ = 1716.
At least 4 girls = ⁴C₄×⁹C₃ = 84. At most 3 = 1716 – 84 = 1632.
Permutations of EXAMINATION in dictionary order. How many words before first word starting with E?
Answer
EXAMINATION: E=1, X=1, A=2, M=1, I=2, N=2, T=1, O=1 (11 letters)
Letters before E alphabetically: A. Words starting with A: 10!/(2!2!1!1!1!1!) = 10!/(2!2!) = 907200
Answer: 907200 words before the first word starting with E.
6-digit numbers from 0, 1, 3, 5, 7, 9 (no repetition) divisible by 10.
Answer
Divisible by 10
⇒ units digit = 0. Remaining 5 places from {1, 3, 5, 7, 9}: 5! = 120
Words with 2 different vowels and 2 different consonants from English alphabet (5 vowels, 21 consonants).
Answer
⁵C₂ × ²¹C₂ × 4! = 10 × 210 × 24 = 50400
12-question paper (5 in Part I, 7 in Part II). Attempt 8 with at least 3 from each part.
Answer
(3 from I, 5 from II): ⁵C₃×⁷C₅ = 10×21 = 210
(4 from I, 4 from II): ⁵C₄×⁷C₄ = 5×35 = 175
(5 from I, 3 from II): ⁵C₅×⁷C₃ = 1×35 = 35
Total = 210 + 175 + 35 = 420
5-card combinations from 52 with exactly one king.
Answer
⁴C₁ × ⁴⁸C₄ = 4 × 194580 = 778320
Seat 5 men and 4 women in a row so women occupy even places.
Answer
Even places: 2, 4, 6, 8 (4 places). Women in even places: 4! ways. Men in 5 odd places: 5! ways.
Total = 4! × 5! = 24 × 120 = 2880
From 25 students, choose 10 for excursion. 3 students will either all join or none join.
Answer
Case 1: All 3 join. Choose 7 from remaining 22: ²²C₇ = 170544
Case 2: None join. Choose 10 from remaining 22: ²²C₁₀ = 646646
Total = 170544 + 646646 = 817190
Letters of ASSASSINATION with all S's together.
Answer
ASSASSINATION: A=3, S=4, I=2, N=2, T=1, O=1 (13 letters)
Treat 4 S's as one unit: 10 objects with A(3), I(2), N(2), T(1), O(1) = 10!/(3!2!2!) = 151200
Important Extra Questions
In how many ways can 7 students be seated in a row if 2 particular students must sit together?
Answer
Treat the 2 together as one unit: 6 units
⇒ 6! arrangements. The 2 can interchange: 2!.
Total = 6! × 2! = 720 × 2 = 1440
How many 4-digit numbers greater than 5000 can be formed from 1, 2, 3, 4, 5, 6 (no repetition)?
Answer
Thousands digit ≥ 5: choices are 5 or 6 (2 choices). Remaining 3 from 5: ⁵P₃ = 60.
Total = 2 × 60 = 120
In how many ways can 5 boys and 5 girls sit in a circle alternately?
Answer
Fix one boy: 4! ways for boys. 5 girls in 5 alternating positions: 5!.
Total = 4! × 5! = 24 × 120 = 2880
Find ⁿCᵣ + ⁿCᵣ₋₁ and show it equals ⁽ⁿ⁺¹⁾Cᵣ.
Answer
ⁿCᵣ + ⁿCᵣ₋₁ = n!/(r!(n–r)!) + n!/((r–1)!(n–r+1)!)
= n!/((r–1)!(n–r)!) × [1/r + 1/(n–r+1)] = n!/((r–1)!(n–r)!) × (n+1)/(r(n–r+1))
= (n+1)!/(r!(n–r+1)!) = ⁽ⁿ⁺¹⁾Cᵣ ∎
How many triangles can be formed by joining vertices of a hexagon?
Answer
Any 3 of 6 vertices form a triangle: ⁶C₃ = 20