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Learn/ 11th Class Mathematics/ Unit 7 /Numericals

Unit 7: Permutations and Combinations — Exercise 7 4

11th Class Mathematics · Unit 7: Permutations and Combinations

1.Evaluate the following: (i) 50C50 (ii) 1000C0 (iii) 10C7 (iv) 20C17
Formula
(i) Combination formula {}nCn = n!n!(n-n)!
Working
Substituting {}50C50 = 50!50!(50-50)!
= 50!50! × 0! = 10!
Result
{}50C50 = 11 = 1
Formula
(ii) Combination formula {}nC0 = n!0!(n-0)!
Working
Substituting {}1000C0 = 1000!0!(1000-0)!
= 1000!1 × 1000!
Result
{}1000C0 = 1
Formula
(iii) Combination formula {}10C7 = 10!7!(10-7)!
= 10.9.8.7!7!3! = 10.9.83.2.1
Result
{}10C7 = 120
Formula
(iv) Combination formula {}20C17 = 20!17!(20-17)!
= 20.19.18.17!17!3! = 20.19.183.2.1
Result
{}20C17 = 1140
2.(i) If 3nC2 : nC2 = 15:1, find n. (ii) If nPr = 120 and nCr = 20, find r.
Given
(i) Ratio {}3nC2 : {}nC2 = 15:1 ⇒ {}3nC2 = 15 × {}nC2
Working
(3n)!2!(3n-2)! = 15n!2!(n-2)!
(3n)(3n-1)(3n-2)!(3n-2)! = 15n(n-1)(n-2)!(n-2)! ⇒ 3n(3n-1) = 15n(n-1)
(3n-1) = 5(n-1) ⇒ 3n-1 = 5n-5
-1+5 = 5n-3n ⇒ 2n = 4
Result
n = 2
Given
(ii) {}nPr = 120, quad {}nCr = 20
Formula
Relation between P and C {}nCr · r! = {}nPr
Working
20 r! = 120 ⇒ r! = 12020
r! = 6 ⇒ r! = 3!
Result
r = 3
3.(i) Find the values of n and r, when nCr = 56, nPr = 336 (ii) Find the values of n and r, when n-1Cr-1 : nCr : n+1Cr+1 = 1:3:7
Given
(i) {}nCr = 56, quad {}nPr = 336
Formula
r! · {}nCr = {}nPr
Working
r!(56) = 336 ⇒ r! = 33656
r! = 6 ⇒ r! = 3! ⇒ r = 3
Consider nPr = 336, put r=3 {}nP3 = 336 ⇒ n!(n-3)! = 336
n(n-1)(n-2) = 336 = 8.7.6
Result
n = 8
Given
(ii) {}n-1Cr-1 : {}nCr : {}n+1Cr+1 = 1:3:7
From first ratio {}n-1Cr-1 : {}nCr = 1:3 ⇒ {}nCr = 3 {}n-1Cr-1
n!r!(n-r)! = 3(n-1)!(r-1)!(n-r)! ⇒ nr = 3 ⇒ n = 3r
From second ratio {}nCr : {}n+1Cr+1 = 3:7 ⇒ 3 {}n+1Cr+1 = 7 {}nCr
3(n+1)r+1 = 7 ⇒ 3(n+1) = 7(r+1)
Working
Putting n = 3r 3(3r+1) = 7(r+1) ⇒ 9r+3 = 7r+7
2r = 4 ⇒ r = 2
Result
n = 3r = 3(2) = 6
4.(i) Prove that nCr + nCr-1 = n+1Cr (ii) Prove that r.nCr = (n-r+1) nCr-1
Given
(i) L.H.S. {}nCr + {}nCr-1 = n!r!(n-r)! + n!(r-1)!(n-r+1)!
= n!(r-1)!(n-r)!left[1r+1n-r+1right] = n!(r-1)!(n-r)! · n+1r(n-r+1)
= (n+1)n!r!(n-r+1)! = (n+1)!r!((n+1)-r)!
Result
= {}n+1Cr = R.H.S.
Given
(ii) R.H.S. (n-r+1) {}nCr-1 = (n-r+1)n!(r-1)!(n-r+1)!
= (n-r+1)n!(r-1)!(n-r+1)(n-r)! = n!(r-1)!(n-r)!
= r.n!r(r-1)!(n-r)! = rn!r!(n-r)!
Result
= r {}nCr = L.H.S.
5.Prove that product of r consecutive integers is divisible by r!.
Given
Let n,n+1,ldots,n+r-1 be r consecutive integers
Product p = (n+r-1)cdots(n+2)(n+1)(n)
In factorial form p = (n+r-1)!(n-1)!
p = (n+r-1)!(n-1)!r! · r! = {}n+r-1Cr · r!
Result
Since {}n+r-1Cr is an integer, p is divisible by r!
6.In how many ways can five subjects be selected out of eight subjects to select a course programme?
Given
Total subjects = 8, quad subjects to select = 5
Formula
Required number of ways {}8C5 = 8!5!(8-5)!
= 8.7.6.5!5!3! = 8.7.63.2.1
Result
{}8C5 = 56
7.Find the number of possible arrangements of vowel letters from the English alphabet?
Given
Total English alphabets = 26, quad vowels to choose = 5
Formula
Required number of ways {}26C5 = 26!5!(26-5)!
= 26.25.24.23.22.21!5!21! = 26.25.24.23.225.4.3.2.1
Result
= 65780
8.In how many ways 3 dishes of Desi foods and 2 dishes of Chinese foods be selected from 6 dishes of desi foods and 8 dishes of Chinese foods?
Given
Desi dishes = 6, to choose = 3; quad Chinese dishes = 8, to choose = 2
Formula
Required number of ways {}6C3 · {}8C2 = 6!3!(6-3)! · 8!2!(8-2)!
= 6.5.43.2.1 · 8.72.1
Result
= 20 × 28 = 560
9.From a standard deck of 52 playing cards, there are 26 black cards and 26 red cards. How many different ways can eight cards be selected if 3 are black and the remaining 5 are red?
Given
Black cards = 26, to select = 3; quad Red cards = 26, to select = 5
Formula
Required number of ways {}26C3 · {}26C5 = 26!3!(26-3)! · 26!5!(26-5)!
= 26.25.243.2.1 · 26.25.24.23.225.4.3.2.1
Result
= 2600 × 65780 = 171028000
10.A bag contains 8 red balls and 7 green balls. Find the total number of possible ways in which five balls are selected in a way: (i) 3 red and 2 green balls (ii) 1 red and 4 green balls (iii) 4 red and 1 green balls (iv) all the red balls
Given
Red balls = 8, quad Green balls = 7
Formula
(i) 3 red and 2 green {}8C3 · {}7C2 = 8.7.63.2.1 · 7.62.1 = 56 × 21 = 1176
(ii) 1 red and 4 green {}8C1 · {}7C4 = 8 · 7.6.53.2.1 = 8 × 35 = 280
(iii) 4 red and 1 green {}8C4 · {}7C1 = 8.7.6.54.3.2.1 · 7 = 70 × 7 = 490
(iv) all the red balls {}8C5 = 8.7.63.2.1 = 56
(i) 1176,quad (ii) 280,quad (iii) 490,quad (iv) 56
11.How many diagonals and triangles can be formed by joining the vertices of the polygon having 15 sides?
Given
Sides = 15, quad Vertices = 15
Formula
Number of diagonals {}15C2 - 15 = 15.142.1 - 15 = 105 - 15 = 90
Number of triangles {}15C3 = 15.14.133.2.1 = 455
Diagonals = 90, quad Triangles = 455
12.Find the number of sides of a polygon if the number of its diagonals is 104.
Given
Number of diagonals = 104
Formula
Number of diagonals = n2-n2 - n = n2-3n2
Working
104 = n2-3n2 ⇒ n2-3n-208 = 0
n2-16n+13n-208=0 ⇒ (n-16)(n+13)=0
Result
n = 16 (n=-13 rejected since nin N)
13.How many triangles can be formed by joining 15 points, 6 of which lie on the same straight line?
Given
Total points = 15, quad Collinear points = 6
Formula
Number of triangles {}15C3 - {}6C3
= 15.14.133.2.1 - 6.5.43.2.1 = 455 - 20
Result
= 435
14.The members of a club are 10 boys and 8 girls. In how many ways can a committee of 6 boys and 3 girls be formed?
Given
Boys = 10, to select = 6; quad Girls = 8, to select = 3
Formula
Required number of ways {}10C6 · {}8C3 = 10.9.8.74.3.2.1 · 8.7.63.2.1
Result
= 210 × 56 = 11760
15.How many committees of 7 members can be chosen from a group of 10 persons when each committee must include 2 particular persons?
Given
2 particular persons included ⇒ remaining 5 chosen from remaining 8 persons
Formula
Required number of committees {}8C5 = 8.7.63.2.1
Result
= 56
16.In how many ways can a cricket team of 11 players be selected out of 17 players? How many of them will include a particular player?
Given
Total players = 17, quad to select = 11
Formula
Number of teams {}17C11 = 17.16.15.14.13.126.5.4.3.2.1
Result
{}17C11 = 12376
Formula
If a particular player is included {}17-1C11-1 = {}16C10 = 16.15.14.13.12.116.5.4.3.2.1
Result
{}16C10 = 8008
17.There are 6 men and 8 women members of a club. How many committees of seven can be formed? (i) with 3 women (ii) with at most 3 women (iii) with at least 5 women
Given
Men = 6, quad Women = 8, quad Committee size = 7
Formula
(i) With 3 women (4 men) {}8C3 · {}6C4 = 56 × 15 = 840
(ii) At most 3 women: possibilities (1W,6M),(2W,5M),(3W,4M) {}8C1 · {}6C6 + {}8C2 · {}6C5 + {}8C3 · {}6C4
Result
= 8 + 168 + 840 = 1016
(iii) At least 5 women: possibilities (5W,2M),(6W,1M),(7W,0M) {}8C5 · {}6C2 + {}8C6 · {}6C1 + {}8C7 · {}6C0
= 840 + 168 + 8 = 1016
18.There are three sections in a question paper; each section has 3 questions. A student has to solve all 5 questions, choosing at least one question from each section. In how many ways can the student make his choice?
Given
Sections = 3, quad questions per section = 3, quad to select = 5
Case I: 3 from one section, 1 each from other two {}3C1 · {}3C3 · {}3C1 · {}3C1 = 3(1)(3)(3) = 27
Case II: 1 from one section, 2 each from other two {}3C1 · {}3C2 · {}3C2 · {}3C1 = 3(3)(3)(3) = 81
Result
Required number of ways 27 + 81 = 108
19.Consider a cryptographic system that generates an 8-character password. Each character in the password can be either a lowercase letter (a-f) or a digit (0-5). How many passwords can be generated if each password must contain exactly 5 lowercase letters and 3 digits: (a) With repetition allowed (b) Without repetition.
20.On Defense Day, Teacher I compiles a list of 10 distinct national songs, while Teacher II prepares a separate list of 10 different national songs (with no overlap between the two lists). The principal needs to select 3 songs from Teacher I's list, and 3 songs from Teacher II's list. Determine the number of possible selection methods when: (i) The order/sequence of the selected songs is important (ii) The order/sequence of the selected songs is not important.
Given
Teacher I: 10 songs, select 3; quad Teacher II: 10 songs, select 3
Formula
(i) Order important {}10P3 · {}10P3 = 10!7! · 10!7! = 720 × 720
Result
= 518400
Formula
(ii) Order not important {}10C3 · {}10C3 = 10.9.83.2.1 · 10.9.83.2.1 = 120 × 120
Result
= 14400