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RD Chapter 17- Combinations Ex-17.2 Interview Questions Answers

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 × 20C=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


Selected

 

RD Chapter 17- Combinations Ex-17.2 Contributors

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