RD Chapter 17 Combinations Ex 17.2 Solutions
Question - 1 : - From a group of 15 cricket players, a team of 11 players is to be chosen. In how many ways can this be done?
Answer - 1 : -
Given:
Number of players = 15
Number of players tobe selected = 11
By using the formula,
nCr = n!/r!(n – r)!
15C11 = 15! / 11! (15 – 11)!
= 15! / (11! 4!)
= [15×14×13×12×11!] /(11! 4!)
= [15×14×13×12] /(4×3×2×1)
= 15×7×13
= 1365
∴ The total numberof ways of choosing 11 players out of 15 is 1365 ways.
Question - 2 : - How many different boat parties of 8, consisting of 5 boys and 3 girls, can be made from 25 boys and 10 girls?
Answer - 2 : -
Given:
Total boys are = 25
Total girls are = 10
Boat party of 8 to bemade from 25 boys and 10 girls, by selecting 5 boys and 3 girls.
So,
By using the formula,
nCr = n!/r!(n – r)!
25C5 × 10C3 =25!/5!(25 – 5)! × 10!/3!(10-3)!
= 25! / (5! 20!) ×10!/(3! 7!)
=[25×24×23×22×21×20!]/(5! 20!) × [10×9×8×7!]/(7! 3!)
= [25×24×23×22×21]/5!× [10×9×8]/(3!)
=[25×24×23×22×21]/(5×4×3×2×1) × [10×9×8]/(3×2×1)
= 5×2×23×11×21 × 5×3×8
= 53130 × 120
= 6375600
∴ The total numberof different boat parties is 6375600 ways.
Question - 3 : - In how many ways can a student choose 5 courses out of 9 courses if 2 courses are compulsory for every student?
Answer - 3 : -
Given:
Total number ofcourses is 9
So out of 9 courses 2courses are compulsory. Student can choose from 7(i.e., 5+2) courses only.
That too out of 5courses student has to choose, 2 courses are compulsory.
So they have to choose3 courses out of 7 courses.
This can be donein 7C3 ways.
By using the formula,
nCr = n!/r!(n – r)!
7C3 = 7! / 3! (7 – 3)!
= 7! / (3! 4!)
= [7×6×5×4!] / (3! 4!)
= [7×6×5] / (3×2×1)
= 7×5
= 35
∴ The total numberof ways of choosing 5 subjects out of 9 subjects in which 2 are compulsory is35 ways.
Question - 4 : - In how many ways can a football team of 11 players be selected from 16 players? How many of these will (i) Include 2 particular players? (ii) Exclude 2 particular players?
Answer - 4 : -
Given:
Total number ofplayers = 16
Number of players tobe selected = 11
So, the combinationis 16C11
By using the formula,
nCr = n!/r!(n – r)!
16C11 = 16! / 11! (16 – 11)!
= 16! / (11! 5!)
= [16×15×14×13×12×11!]/ (11! 5!)
= [16×15×14×13×12] /(5×4×3×2×1)
= 4×7×13×12
= 4368
(i) Include 2particular players?
It is told that twoplayers are always included.
Now, we have to select9 players out of the remaining 14 players as 2 players are already selected.
Number of ways = 14C9
14C9 = 14! / 9! (14 – 9)!
= 14! / (9! 5!)
= [14×13×12×11×10×9!]/ (9! 5!)
= [14×13×12×11×10] /(5×4×3×2×1)
= 7×13×11×2
= 2002
(ii) Exclude 2particular players?
It is told that twoplayers are always excluded.
Now, we have to select11 players out of the remaining 14 players as 2 players are already removed.
Number of ways = 14C9
14C11 = 14! / 11! (14 – 11)!
= 14! / (11! 3!)
= [14×13×12×11!] /(11! 3!)
= [14×13×12] / (3×2×1)
= 14×13×2
= 364
∴ The required no.of ways are 4368, 2002, 364.
Question - 5 : - There are 10 professors and 20 students out of whom a committee of 2 professors and 3 students is to be formed. Find the number of ways in which this can be done. Further, find in how many of these committees:
(i) a particular professor is included.
(ii) a particular student is included.
(iii) a particular student is excluded.
Answer - 5 : -
Given:
Total number ofprofessor = 10
Total number ofstudents = 20
Number of ways =(choosing 2 professors out of 10 professors) × (choosing 3 students out of 20students)
= (10C2)× (20C3)
By using the formula,
nCr = n!/r!(n – r)!
10C2 × 20C3 =10!/2!(10 – 2)! × 20!/3!(20-3)!
= 10!/(2! 8!) ×20!/(3! 17!)
= [10×9×8!]/(2! 8!) ×[20×19×18×17!]/(17! 3!)
= [10×9]/2! ×[20×19×18]/(3!)
= [10×9]/(2×1) ×[20×19×18]/(3×2×1)
= 5×9 × 10×19×6
= 45 × 1140
= 51300 ways
(i) a particular professoris included.
Number of ways =(choosing 1 professor out of 9 professors) × (choosing 3 students out of 20students)
= 9C1 × 20C3
By using the formula,
nCr = n!/r!(n – r)!
9C1 × 20C3 =9!/1!(9 – 1)! × 20!/3!(20-3)!
= 9!/(1! 8!) × 20!/(3!17!)
= [9×8!]/(8!) ×[20×19×18×17!]/(17! 3!)
= 9 × [20×19×18]/(3!)
= 9×[20×19×18]/(3×2×1)
= 9 × 10×19×6
= 10260 ways
(ii) a particular studentis included.
Number of ways =(choosing 2 professors out of 10 professors) × (choosing 2 students out of 19students)
= 10C2 × 19C2
By using the formula,
nCr = n!/r!(n – r)!
10C2 × 19C2 =10!/2!(10 – 2)! × 19!/2!(19-2)!
= 10!/(2! 8!) ×19!/(2! 17!)
= [10×9×8!]/(2! 8!) ×[19×18×17!]/(17! 2!)
= [10×9]/2! ×[19×18]/(2!)
= [10×9]/(2×1) ×[19×18]/(2×1)
= 5×9 × 19×9
= 45 × 171
= 7695 ways
(iii) a particular studentis excluded.
Number of ways =(choosing 2 professors out of 10 professors) × (choosing 3 students out of 19students)
= 10C2 × 19C3
By using the formula,
nCr = n!/r!(n – r)!
10C2 × 19C3 =10!/2!(10 – 2)! × 19!/3!(19-3)!
= 10!/(2! 8!) ×19!/(3! 16!)
= [10×9×8!]/(2! 8!) ×[19×18×17×16!]/(16! 3!)
= [10×9]/2! ×[19×18×17]/(3!)
= [10×9]/(2×1) ×[19×18×17]/(3×2×1)
= 5×9 × 19×3×17
= 45 × 969
= 43605 ways
∴ The required no.of ways are 51300, 10260, 7695, 43605.
Question - 6 : - How many different products can be obtained by multiplying two or more of the numbers 3, 5, 7, 11 (without repetition)?
Answer - 6 : -
Given that we need tofind the no. of ways of obtaining a product by multiplying two or more from thenumbers 3, 5, 7, 11.
Number of ways = (no.of ways of multiplying two numbers) + (no. of ways of multiplying threenumbers) + (no. of multiplying four numbers)
= 4C2 + 4C3 + 4C4
By using the formula,
nCr = n!/r!(n – r)!
= 12/2 + 4 + 1
= 6 + 4 + 1
= 11
∴ The total numberof ways of product is 11 ways.
Question - 7 : - From a class of 12 boys and 10 girls, 10 students are to be chosen for the competition, at least including 4 boys and 4 girls. The 2 girls who won the prizes last year should be included. In how many ways can the selection be made?
Answer - 7 : -
Given:
Total number of boys =12
Total number of girls= 10
Total number of girlsfor the competition = 10 + 2 = 12
Number of ways = (no.of ways of selecting 6 boys and 2 girls from remaining 12 boys and 8 girls) + (no.of ways of selecting 5 boys and 3 girls from remaining 12 boys and 8 girls) +(no. of ways of selecting 4 boys and 4 girls from remaining 12 boys and 8girls)
Since, two girls arealready selected,
= (12C6 × 8C2)+ (12C5 × 8C3) + (12C4 × 8C4)
By using the formula,
nCr = n!/r!(n – r)!
= (924 × 28) + (792 ×56) + (495 × 70)
= 25872 + 44352 +34650
= 104874
∴ The total numberof ways of product is 104874 ways.
Question - 8 : - How many different selections of 4 books can be made from 10 different books, if
(i) there is no restriction
(ii) two particular books are always selected
(iii) two particular books are never selected
Answer - 8 : -
Given:
Total number of books= 10
Total books to beselected = 4
(i) there is norestriction
Number of ways =choosing 4 books out of 10 books
= 10C4
By using the formula,
nCr = n!/r!(n – r)!
10C4 = 10! / 4! (10 – 4)!
= 10! / (4! 6!)
= [10×9×8×7×6!] / (4!6!)
= [10×9×8×7] /(4×3×2×1)
= 10×3×7
= 210 ways
(ii) two particular booksare always selected
Number of ways =select 2 books out of the remaining 8 books as 2 books are already selected.
= 8C2
By using the formula,
nCr = n!/r!(n – r)!
8C2 = 8! / 2! (8 – 2)!
= 8! / (2! 6!)
= [8×7×6!] / (2! 6!)
= [8×7] / (2×1)
= 4×7
= 28 ways
(iii) two particular booksare never selected
Number of ways = select4 books out of remaining 8 books as 2 books are already removed.
= 8C4
By using the formula,
nCr = n!/r!(n – r)!
8C4 = 8! / 4! (8 – 4)!
= 8! / (4! 4!)
= [8×7×6×5×4!] / (4!4!)
= [8×7×6×5] /(4×3×2×1)
= 7×2×5
= 70 ways
∴ The required no.of ways are 210, 28, 70.
Question - 9 : - From 4 officers and 8 jawans in how many ways can 6 be chosen (i) to include exactly one officer (ii) to include at least one officer?
Answer - 9 : -
Given:
Total number ofofficers = 4
Total number of jawans= 8
Total number ofselection to be made is 6
(i) to include exactly oneofficer
Number of ways = (no.of ways of choosing 1 officer from 4 officers) × (no. of ways of choosing 5jawans from 8 jawans)
= (4C1)× (8C5)
By using the formula,
nCr = n!/r!(n – r)!
(ii) to include at leastone officer?
Number of ways =(total no. of ways of choosing 6 persons from all 12 persons) – (no. of ways ofchoosing 6 persons without any officer)
= 12C6 – 8C6
By using the formula,
nCr = n!/r!(n – r)!
= (11×2×3×2×7) – (4×7)
= 924 – 28
= 896 ways
∴ The required no.of ways are 224 and 896.
Question - 10 : - A sports team of 11 students is to be constituted, choosing at least 5 from class XI and at least 5 from class XII. If there are 20 students in each of these classes, in how many ways can the teams be constituted?
Answer - 10 : -
Given:
Total number ofstudents in XI = 20
Total number ofstudents in XII = 20
Total number ofstudents to be selected in a team = 11 (with atleast 5 from class XI and 5 fromclass XII)
Number of ways = (No.of ways of selecting 6 students from class XI and 5 students from class XII) +(No. of ways of selecting 5 students from class XI and 6 students from classXII)
= (20C6 × 20C5)+ (20C5 × 20C6)
= 2 (20C6 × 20C5)ways