StudentAid
HomeLearn
9th Class10th Class11th Class12th ClassCambridge past papersBrowse everything
Attendance
Class attendanceVisitor register
For schoolsFor driversFor parentsBook a demo
Learn/ 11th Class Mathematics/ Unit 7 /Numericals

Unit 7: Permutations and Combinations — Exercise 7 2

11th Class Mathematics · Unit 7: Permutations and Combinations

1.(i) Evaluate ¹⁰P₅ (ii) Evaluate ⁵P₂ (iii) Evaluate ⁷P₇ (iv) Evaluate ¹⁰P₃
Formula
(i) Permutation formula 10P5 = 10!(10-5)!
Working
Expanding = 10 · 9 · 8 · 7 · 6 · 5!5!
= 10 · 9 · 8 · 7 · 6
Result
= 30240
Formula
(ii) Permutation formula 5P2 = 5!(5-2)!
Working
Expanding = 5 · 4 · 3!3!
= 5 × 4
Result
= 20
Formula
(iii) Permutation formula 7P7 = 7!(7-7)!
Working
Expanding = 7 · 6 · 5 · 4 · 3 · 2 · 10!
= 50401
Result
= 5040
Formula
(iv) Permutation formula 10P3 = 10!(10-3)!
Working
Expanding = 10 · 9 · 8 · 7!7!
= 10 · 9 · 8
Result
= 720
2.(i) Find the value of n when ⁿP₃ = 504 (ii) Find the value of n when ¹⁵Pₙ = 15.14.13.12.11 (iii) Find the value of n when ⁿP₅ : ⁿ⁻²P₂ = 540 : 1
Given
(i) Given nP3 = 504
Formula
Permutation formula n!(n-3)! = 504
Working
Expanding n(n-1)(n-2) = 504
Testing values n(n-1)(n-2) = 9 · 8 · 7 = 504
Result
n = 9
Given
(ii) Given 15Pn = 15 · 14 · 13 · 12 · 11
Formula
Permutation formula 15!(15-n)! = 15 · 14 · 13 · 12 · 11
Counting factors 15 · 14 · 13 · 12 · 11 · 10!10!
Working
Setting equal 15!(15-n)! = 15!10!
Therefore (15-n)! = 10! ⇒ 15-n = 10
Result
n = 5
Given
(iii) Given nP5 : n-2P2 = 540 : 1
Formula
Ratio of permutations nP5n-2P2 = 540
Working
Expanding n!/(n-5)!(n-2)!/(n-4)! = 540
Simplifying n(n-1)(n-2)!(n-5)! · (n-4)!(n-2)! = 540
Canceling n(n-1)(n-4)!(n-5)! = 540
Further simplifying n(n-1)(n-4) = 540
Testing 10 · 9 · 6 = 540
Result
n = 10
4.(i) How many words can be formed from the letters of the word PYTHON using all letters when no letter is to be repeated? (ii) How many words can be formed from the letters of the word NETWORK using all letters when no letter is to be repeated? (iii) How many words can be formed from the letters of the word COMPUTER using all letters when no letter is to be repeated?
Given
(i) Total letters n = 6
Letters to use r = 6
Formula
Permutations 6P6 = 6!
Calculating = 6 · 5 · 4 · 3 · 2 · 1
Result
= 720
Given
(ii) Total letters n = 7
Letters to use r = 7
Formula
Permutations 7P7 = 7!
Calculating = 7 · 6 · 5 · 4 · 3 · 2 · 1
Result
= 5040
Given
(iii) Total letters n = 8
Letters to use r = 8
Formula
Permutations 8P8 = 8!
Calculating = 8 · 7 · 6 · 5 · 4 · 3 · 2 · 1
Result
= 40320
5.How many signals can be given by 6 flags of different colours, using 2 flags at a time?
Given
Number of flags n = 6
Flags used at a time r = 2
Formula
Permutations 6P2 = 6!(6-2)!
Working
Calculating = 6 · 5 · 4!4!
= 6 × 5
Result
= 30
6.How many signals can be given by 5 flags of different colours, when any number of flags are used at a time?
7.How many 4-digit numbers can be formed, with distinct digits, with each digit odd?
Given
Odd digits 1, 3, 5, 7, 9
Total odd digits n = 5
Digits to select r = 4
Formula
Permutations 5P4 = 5!(5-4)!
Working
Calculating = 5 · 4 · 3 · 2 · 1!1!
= 5 · 4 · 3 · 2
Result
= 120
8.How many numbers between 100 and 1000 can be formed by using the digits 0,1,2,3,4,5 without repetition? How many of them are divisible by 5?
Given
Given digits 0, 1, 2, 3, 4, 5
Total digits n = 6
Formula
3-digit numbers without restriction 6P3 = 6!(6-3)! = 6 · 5 · 4 · 3!3! = 120
Numbers with 0 at units place squaresquare 0 = 5 × 4 = 20
Result
Numbers between 100 and 1000 = 120 - 20 = 100
Numbers divisible by 5 (ending in 0) squaresquare 0 = 5 × 4 = 20
Numbers divisible by 5 (ending in 5) squaresquare 5 = 4 × 4 = 16
Total divisible by 5 = 20 + 16 = 36
9.Find the numbers greater than 35000 that can be formed from the digits 1,2,3,4,5,6, without repeating any digit.
Form 35____ 4P4 = 4!0! = 24
Form 36____ 4P3 = 4!(4-3)! = 24
Form 4____ 5P4 = 5!(5-4)! = 5!1! = 120
Form 5____ 5P4 = 120
Form 6____ 5P4 = 120
Result
Total = 24 + 24 + 120 + 120 + 120 = 408
10.(i) Find the number of 5-digit numbers that can be formed from the digits 1,2,4,6,8 (when no digit is repeated) such that the digits 2 and 8 are next to each other. (ii) Find the number of 5-digit numbers that can be formed from the digits 1,2,4,6,8 (when no digit is repeated) such that the digits 2 and 8 are NOT next to each other.
(i) Treating 2 and 8 as one unit Consider 28 and 82 as single entities
For arrangement 28 4P4 = 4! = 24
For arrangement 82 4P4 = 4! = 24
Result
= 24 + 24 = 48
Given
(ii) Total permutations of 5 digits 5P5 = 5! = 120
Permutations with 2 and 8 together = 48
Result
Permutations with 2 and 8 NOT together = 120 - 48 = 72
11.How many 6-digit numbers can be formed without repeating any digit from the digits 0,1,2,3,4,5? In how many of them will 0 be at the tens place?
Given
Available digits 0, 1, 2, 3, 4, 5
First digit options (not 0) 5P1 = 5 ways
Remaining 5 places with remaining 5 digits 5P5 = 5! = 120 ways
Result
Total 6-digit numbers = 5 × 120 = 600
With 0 at tens place: remaining 5 digits in 5 places 5P5 = 5! = 120
12.How many 5-digit multiples of 5 can be formed from the digits 2,3,5,7,9 when no digit is repeated.
Given
Available digits 2, 3, 5, 7, 9
For multiple of 5, fix 5 at units place Last digit = 5
Remaining 4 digits in 4 places 4P4 = 4! = 24
Result
= 24
13.In how many ways can 8 different books including 2 on English be arranged on a shelf in such a way that the English books are never together.
Given
Total books n = 8
English books r = 2
Total permutations of 8 books 8P8 = 8! = 40320
Permutations with English books together (E₁E₂ or E₂E₁) 7P1 × 2P2 = 7! + 7! = 5040 + 5040 = 10080
Result
English books NOT together = 40320 - 10080 = 30240
14.Find the number of arrangements of 3 different books on English and 5 different books on Urdu for placing them on a shelf such that the books on the same subject are together.
Given
English books n1 = 3
Urdu books n2 = 5
English first, then Urdu 3P3 × 5P5 = 3! × 5! = 6 × 120 = 720
Urdu first, then English 5P5 × 3P3 = 5! × 3! = 120 × 6 = 720
Result
Total arrangements = 720 + 720 = 1440
15.In how many ways can 5 boys and 4 girls be seated on a bench so that the girls and the boys occupy alternate seats?
Given
Number of boys m = 5
Number of girls n = 4
Seating arrangement B1 G1 B2 G2 B3 G3 B4 G4 B5
Arrange 5 boys in 5 seats 5P5 = 5! = 120
Arrange 4 girls in 4 seats 4P4 = 4! = 24
Result
Total ways = 120 × 24 = 2880