Question 1
How many 3-digit numbers can be made from 1,2,3,4,5 if (i) repetition is allowed; (ii) repetition is not allowed?
Steps only
With repetition: 5 choices in each of 3 places, so 5³=125. Without repetition: 5×4×3=60.
Answer
- (i) 125;
- (ii) 60.
CLASS 11 · MATHEMATICS · CHAPTER 6
Permutations and Combinations
Count ordered arrangements and unordered selections with clear multiplication, permutation and combination reasoning.
42 worked answersAll 4 exercise setsAll 11 end-of-chapter questions
What you will learn
- Use the fundamental principle of counting.
- Calculate permutations with and without repeated objects.
- Distinguish arrangements from selections.
- Apply combinations to teams, cards and committees.
42 worked answers · short steps first, explanations when you need them
Exercise 6.1
Textbook pages 104
Question 2
How many 3-digit even numbers can be formed from 1,2,3,4,5,6 if repetition is allowed?
Steps only
The units digit must be 2,4 or 6 (3 choices). Hundreds and tens each have 6 choices: 6×6×3=108.
Answer
108.
Question 3
How many 4-letter codes use the first 10 English letters if no letter repeats?
Steps only
Choose letters in order: 10×9×8×7=5,040.
Answer
5,040.
Question 4
How many 5-digit telephone numbers start with 67 and use digits 0–9 without repetition?
Steps only
The first two places are fixed. The remaining three places have 8, then 7, then 6 choices: 8×7×6=336.
Answer
336.
Question 5
A coin is tossed three times. How many possible outcome sequences are there?
Steps only
Each toss has 2 outcomes, so 2×2×2=2³=8.
Answer
8.
Question 6
Five distinct flags make a signal using 2 flags, one below the other. How many signals?
Steps only
Order matters because top and bottom differ. Choose the top in 5 ways and bottom in 4: 5×4=20.
Answer
20.
Exercise 6.2
Textbook pages 106–107
Question 1
Evaluate (i) 8!; (ii) 4!−3!.
Steps only
8!=8×7×…×1=40,320. Also 4!−3!=24−6=18.
Answer
- (i) 40,320;
- (ii) 18.
Question 2
Is 3!+4!=7!?
Steps only
3!+4!=6+24=30, while 7!=5,040.
Answer
No: 30≠5,040.
Question 3
Compute 8!/(6!2!).
Steps only
Cancel 6!: (8×7)/(2×1)=28.
Answer
28.
Question 4
If 1/6!+1/7!=x/8!, find x.
Steps only
Multiply through by 8!: x=8!/6!+8!/7!=8×7+8=56+8=64.
Answer
x=64.
Question 5
Evaluate n!/(n−r)! for (i) n=6,r=2; (ii) n=9,r=5.
Steps only
(i) 6!/4!=6×5=30. (ii) 9!/4!=9×8×7×6×5=15,120.
Answer
- (i) 30;
- (ii) 15,120.
Exercise 6.3
Textbook pages 114–115
Question 1
How many 3-digit numbers can be formed from digits 1–9 without repetition?
Steps only
There are 9 choices for the hundreds digit, then 8 and 7: 9×8×7=504.
Answer
504.
Question 2
How many 4-digit numbers can be formed without repeating any digit?
Steps only
The first digit has 9 choices (1–9); the next places have 9,8,7 choices (zero is now allowed): 9×9×8×7=4,536.
Answer
4,536.
Question 3
How many 3-digit even numbers use 1,2,3,4,6,7 without repetition?
Steps only
Choose the units digit from {2,4,6}: 3 choices. Then hundreds and tens have 5 and 4 choices: 3×5×4=60.
Answer
60.
Question 4
From digits 1,2,3,4,5 without repetition, how many 4-digit numbers are possible, and how many are even?
Steps only
Total: 5P4=5×4×3×2=120. For even numbers, last digit has 2 choices (2 or 4), and the first three places have 4×3×2 choices: 48.
Answer
Total 120; even 48.
Question 5
From 8 people choose a chairman and vice-chairman, with no one holding both offices. How many ways?
Steps only
The roles are different: 8 choices for chair and 7 for vice-chair, so 8×7=56.
Answer
56.
Question 6
Find n if (n−1)P3 : nP4 = 1:9.
Steps only
(n−1)P3/nP4 = [(n−1)(n−2)(n−3)]/[n(n−1)(n−2)(n−3)]=1/n. Thus 1/n=1/9 and n=9.
Answer
n=9.
Question 7
Solve for r: (i) ⁵Pᵣ = 2·⁶Pᵣ₋₁; (ii) ⁵Pᵣ = ⁶Pᵣ₋₁.
Steps only
Use ⁿPᵣ=n!/(n−r)!. (i) 5!/(5−r)! = 2·6!/(7−r)!, so (7−r)(6−r)=12. This gives r²−13r+30=0, or r=3 or 10; since r≤5, r=3.
(ii) 5!/(5−r)! = 6!/(7−r)!, so (7−r)(6−r)=6. This gives r²−13r+36=0, or r=4 or 9; since r≤5, r=4.
Answer
- (i) r=3;
- (ii) r=4.
Question 8
How many arrangements use all letters of EQUATION exactly once?
Steps only
The 8 letters are distinct, so there are 8!=40,320 arrangements.
Answer
40,320.
Question 9
For MONDAY with no repeated letter, find (i) 4-letter words; (ii) arrangements of all letters; (iii) arrangements of all letters beginning with a vowel.
Steps only
(i) 6P4=360. (ii) 6!=720. (iii) The first letter has 2 choices (A or O); arrange the other 5 in 5! ways: 2×120=240.
Answer
- (i) 360;
- (ii) 720;
- (iii) 240.
Question 10
How many distinct arrangements of MISSISSIPPI have the four I’s not all together?
Steps only
There are 11 letters with I⁴,S⁴,P²,M: total 11!/(4!4!2!)=34,650. Treating all four I’s as one block gives 8!/(4!2!)=840. Subtract: 34,650−840=33,810.
Answer
33,810.
Question 11
For PERMUTATIONS (two T’s), find arrangements (i) beginning P and ending S; (ii) with vowels together; (iii) with exactly four letters between P and S.
Steps only
(i) Arrange the remaining 10 letters: 10!/2!. (ii) Treat five vowels as a block: 8!/2!×5!. (iii) Choose one of 7 endpoint pairs and their order (14 choices); arrange the other 10 letters in 10!/2! ways.
Answer
- (i) 1,814,400;
- (ii) 2,419,200;
- (iii) 25,401,600.
Exercise 6.4
Textbook pages 119
Question 1
If nC8=nC2, find nC2.
Steps only
For the same n, nCr=nC(n−r). Equality with r=2 and 8 gives n−2=8, hence n=10; 10C2=45.
Answer
n=10, so nC2=45.
Question 2
Find n if (i) 2nC3:nC3=12:1; (ii) 2nC3:nC3=11:1.
Steps only
The ratio is (2nC3)/(nC3)=4(2n−1)/(n−2). Set this equal to 12 or 11 and solve, keeping n≥3.
Answer
- (i) n=5;
- (ii) n=6.
Question 3
How many chords can be drawn from 21 points on a circle?
Steps only
Each chord is determined by choosing 2 endpoints: 21C2=21×20/2=210.
Answer
210.
Question 4
How many teams of 3 boys and 3 girls can be selected from 5 boys and 4 girls?
Steps only
Choose 3 boys in 5C3=10 ways and 3 girls in 4C3=4 ways. Multiply: 40.
Answer
40.
Question 5
Select 9 balls from 6 red, 5 white and 5 blue, with exactly 3 of each colour.
Steps only
Choose separately: 6C3×5C3×5C3=20×10×10=2,000.
Answer
2,000.
Question 6
How many 5-card hands contain exactly one ace?
Steps only
Choose 1 of 4 aces and 4 of the 48 non-aces: 4×48C4=4×194,580=778,320.
Answer
778,320.
Question 7
Choose an 11-player cricket team from 17, including exactly 4 of the 5 bowlers.
Steps only
Choose 4 bowlers in 5C4=5 ways. Choose the remaining 7 from 12 non-bowlers in 12C7=792 ways. Product=3,960.
Answer
3,960.
Question 8
A bag has 5 black and 6 red balls. In how many ways can 2 black and 3 red be selected?
Steps only
Choose independently: 5C2×6C3=10×20=200.
Answer
200.
Question 9
Choose a 5-course programme from 9 courses when 2 specific courses are compulsory.
Steps only
The two required courses are already selected. Choose 3 more from the remaining 7: 7C3=35.
Answer
35.
Miscellaneous Exercise
Textbook pages 120–123
Question 1
From DAUGHTER, how many 5-letter words contain exactly 2 vowels and 3 consonants?
Steps only
DAUGHTER has 3 vowels and 5 consonants. Choose 2 vowels and 3 consonants, then arrange the five chosen letters: 3C2×5C3×5!=3×10×120=3,600.
Answer
3,600.
Question 2
Arrange EQUATION so that all vowels occur together and all consonants occur together.
Steps only
Treat the five vowels as one block and three consonants as another. Arrange the two blocks in 2! ways; arrange within them in 5! and 3! ways: 2×120×6=1,440.
Answer
1,440.
Question 3
Form a 7-person committee from 9 boys and 4 girls with (i) exactly 3 girls; (ii) at least 3 girls; (iii) at most 3 girls.
Steps only
(i) 4C3×9C4=504. (ii) Add the 4-girl case: 504+9C3=588. (iii) All committees total 13C7=1,716; subtract the 4-girl case 9C3=84, giving 1,632.
Answer
- (i) 504;
- (ii) 588;
- (iii) 1,632.
Question 4
In dictionary order, how many distinct arrangements of EXAMINATION come before the first one beginning with E?
Steps only
Only A is alphabetically before E among the first letters. Fix A first; the remaining 10 letters have I repeated twice and N repeated twice, so 10!/(2!2!)=907,200.
Answer
907,200.
Question 5
How many 6-digit numbers use 0,1,3,5,7,9 once each and are divisible by 10?
Steps only
A number divisible by 10 must end in 0. Arrange the other five digits freely: 5!=120.
Answer
120.
Question 6
How many 4-letter words have two distinct vowels and two distinct consonants from the English alphabet?
Steps only
Choose 2 vowels in 5C2 ways and 2 consonants in 21C2 ways; arrange the 4 chosen letters in 4! ways: 10×210×24=50,400.
Answer
50,400.
Question 7
Choose 8 questions from Part I (5 questions) and Part II (7 questions), taking at least 3 from each.
Steps only
The allowed splits are (3,5),(4,4),(5,3). Add 5C3×7C5 + 5C4×7C4 + 5C5×7C3 = 210+175+35=420.
Answer
420 ways.
Question 8
How many 5-card hands contain exactly one king?
Steps only
Choose 1 of 4 kings and 4 of the 48 other cards: 4×48C4=778,320.
Answer
778,320.
Question 9
Seat 5 men and 4 women in a row so the women occupy even-numbered positions.
Steps only
There are four even seats among positions 1–9. Arrange the women in them in 4! ways and the men in the remaining five places in 5! ways: 4!×5!=2,880.
Answer
2,880.
Question 10
Choose 10 students from 25 when 3 named students either all join or none join.
Steps only
If all three join, choose 7 of the other 22: 22C7=170,544. If none join, choose all 10 from 22: 22C10=646,646. Add to get 817,190.
Answer
817,190 ways.
Question 11
Arrange ASSASSINATION so all four S’s stay together.
Steps only
Treat SSSS as one block. There are 10 objects total with A repeated 3 times, I repeated 2 times and N repeated 2 times. Count 10!/(3!2!2!)=151,200.
Answer
151,200.