Combination Function and its properties MCQ Quiz - Objective Question with Answer for Combination Function and its properties - Download Free PDF

Last updated on Jun 19, 2025

Latest Combination Function and its properties MCQ Objective Questions

Combination Function and its properties Question 1:

Comprehension:

Direction: We have to choose Eleven players for cricket team from eight batsmen, six bowlers, four all rounder and two wicket keepers in the following conditions.

Number of selections when a particular batsman and a particular wicket keeper do not want to play together 

  1. 218C10
  2. 19C11+18C10
  3. 19C10+19C11
  4. None of these

Answer (Detailed Solution Below)

Option 2 : 19C11+18C10

Combination Function and its properties Question 1 Detailed Solution

Calculation:

Case 1: If the particular batsman is selected.

 

Then, the rest of the 10 players can be selected from the remaining 18 players. The number of ways to select these 10 players is:

(1810)

Case 2: If the particular wicket-keeper is selected.

Then, the rest of the 10 players can be selected from the remaining 18 players. The number of ways to select these 10 players is:

(1810)

Case 3: If both the particular batsman and wicket-keeper are not selected.

The number of ways to select 11 players from the remaining 18 players is:

(1811)

Total number of selections:

The total number of selections is:

2(1810)+(1811)=19C11+18C10

Hence, the correct answer is option 2.

Combination Function and its properties Question 2:

Comprehension:

Direction: We have to choose Eleven players for cricket team from eight batsmen, six bowlers, four all rounder and two wicket keepers in the following conditions.

Number of selections, when two particular batsmen do not want to play when a particular bowler will play

  1. 17C10+19C11
  2. 17C10+19C11+17C11
  3. 17C10+20C11
  4. 19C10+19C11

Answer (Detailed Solution Below)

Option 2 : 17C10+19C11+17C11

Combination Function and its properties Question 2 Detailed Solution

Calculation:

The condition is that if the particular bowler plays, then two particular batsmen will not play.

 

Case 1: The particular bowler plays, and the two particular batsmen do not play.

In this case, we are left with 17 other players to choose from (excluding the two batsmen and the particular bowler).

The number of selections is:

(1710)

Case 2: The particular bowler does not play.

If the bowler does not play, we can choose from the remaining 19 players (excluding the particular bowler). We need to select 11 players from these 19 players:

(1911)

Case 3: All three players (the particular bowler and the two particular batsmen) do not play.

If all three players do not play, we are left with 17 players. We need to choose 11 players from these 17:

(1711)

Total number of selections:

Now, the total number of selections is the sum of the selections from all three cases:

(1710)+(1911)+(1711)

Hence, the total number of selections is:

(1710)+(1911)+(1711)

Combination Function and its properties Question 3:

Comprehension:

Direction: We have to choose Eleven players for cricket team from eight batsmen, six bowlers, four all rounder and two wicket keepers in the following conditions.

The number of selections, when at most one all rounder and one wicket keeper will play

  1. 4C1×14C10+2C1×14C10+4C1×2C1×14C9+14C11
  2. 4C1×15C11+15C11
  3. 4C1×15C10+15C11
  4. None of these

Answer (Detailed Solution Below)

Option 1 : 4C1×14C10+2C1×14C10+4C1×2C1×14C9+14C11

Combination Function and its properties Question 3 Detailed Solution

Concept:

The Combination means “Selection of things”, where the order of things has no importance

r = 0 ,  when n < r

Where n = distinct object to choose from

C = Combination

r = spaces to fill 

Calculation:

Total number of players selected for cricket team = 11

Number of batsmen = 8

Number of bowlers = 6

Number of all-rounders = 4

number of wicket-keepers = 2

Arrangements when at most 1 all-rounder and 1 wicket-keeper.

When 1 all-rounder, 1 wicket-keeper, and 9 remaining players play,

Number of ways = 4C1.2C1.14C9

When 0 all-rounders, 1 wicket-keeper, and 10 remaining players play,

Number of ways = 2C1.14C10

When 0 all-rounders, 0 wicket-keeper, and 11 remaining players play,

Number of ways = 14C11 

When 1 all-rounder, 0 wicket-keeper and 10 remaining players play,

Number of ways = 4C1.14C10

∴ Total number of selections = 4C1.2C1.14C+ 2C1.14C10 + 14C11 + 4C1.14C10

Combination Function and its properties Question 4:

What is the number of selections of at most 3 things from 6 different things?

  1. 20
  2. 22
  3. 41
  4. 42

Answer (Detailed Solution Below)

Option 4 : 42

Combination Function and its properties Question 4 Detailed Solution

Calculation:

Given:

Total number of items, n = 6

We need to calculate the total number of selections for r = 0, 1, 2, and 3:

For r = 0

6C0=6!0!(60)!=11=1

⇒ Number of ways to select 0 items = 1

For r = 1

6C1=6!1!(61)!=61=6

⇒ Number of ways to select 1 item = 6

For r = 2

6C2=6!2!(62)!=6×52×1=15

⇒ Number of ways to select 2 items = 15

For r = 3

6C3=6!3!(63)!=6×5×43×2×1=20

⇒ Number of ways to select 3 items = 20

Total number of selections:

6C0+6C1+6C2+6C3=1+6+15+20=42

∴ The total number of selections of at most 3 things from 6 different things is 42.

Hence, the correct answer is Option 4.

Combination Function and its properties Question 5:

What is the maximum number of possible points of intersection of four straight lines and a circle (intersection is between lines as well as circle and lines)? 

  1. 6
  2. 10
  3. 14
  4. 16

Answer (Detailed Solution Below)

Option 3 : 14

Combination Function and its properties Question 5 Detailed Solution

Explanation:

Since there are 4 lines, the maximum number of intersection points are 4C2 = 6.

The maximum number of intersections of the circle with 4 lines is 4 × 2 = 8.

Required intersection points = 6 + 8 = 14

∴ the correct answer is Option 3 ie 14

Top Combination Function and its properties MCQ Objective Questions

If (158)+(157)=(nr)

then the values of n and r are:

Where, (nr)=nCr

  1. 16 and 7.
  2. 16 and 8.
  3. 16 and 9.
  4. 30 and 15.

Answer (Detailed Solution Below)

Option 2 : 16 and 8.

Combination Function and its properties Question 6 Detailed Solution

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Concept:

  • nCr = nCn-r.
  • If nCx = nCy, then x = y or x + y = n.
  • nCr + nCr-1 = n+1Cr.

 

Calculations:

(158)+(157)=(nr)

⇒ 15C8 + 15C7 = nCr

Using nCr + nCr-1 = n+1Cr:

⇒ ​15C8 + 15C8-1 = 15+1C8nCr

⇒ ​16C8 = nCr

n = 16 and r = 8.

If C(20, n + 2) = C(20, n – 2) then what is n equal to?

  1. 8
  2. 10
  3. 12
  4. 16

Answer (Detailed Solution Below)

Option 2 : 10

Combination Function and its properties Question 7 Detailed Solution

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Concept:

The notation C(n, r) is the number of combinations/groups of n different things taking r at a time and is given by: C(n,r)=n!r!(nr)!

If C(n, x) = C(n, y), then x + y = n.

Calculation:

Given C(20, n + 2) = C(20, n – 2)

As we know that, if C(n, x) = C(n, y), then x + y = n.

⇒ (n + 2) + (n – 2) = 20

⇒ 2n = 20

⇒ n = 10

Answer (Detailed Solution Below)

Option 3 : 495

Combination Function and its properties Question 8 Detailed Solution

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Concept:

If nCx=nCy then x + y = n

nCr=n×(n1)×....(nr+1)r!

 

Calculation: 

Given: nCnC7

It is possible only when n = 5 + 7 = 12

∴ nC4=12C4=12×11×10×94!

=12×11×10×94×3×2

= 495

Hence, option (3) is correct.

In an examination a candidate has to pass in each of the 5 subjects. In how many ways he can fail ?

  1. 31
  2. 20
  3. 25
  4. 30

Answer (Detailed Solution Below)

Option 1 : 31

Combination Function and its properties Question 9 Detailed Solution

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Concept:

The number of ways to select r things out of n given things wherein r ≤ n is given by: nCr=n!r!×(nr)!

Calculation:

Given: There are 5 subjects and for a student to pass in an examination the students has to pass in each of the 5 subjects.

Here, we have to find in how many ways a student can fail in the examination.

In order to fail in the examination the student can fail in 1 or 2 or 3 or 4 or 5 subjects out of the 5 subjects in each case.

Case 1: Student fails in any 1 subject out of the 5 subjects.

No. of ways in which student can fail in any 1 subject out of the 5 subjects = 5C1

Case 2: Student fails in any 2 subjects out of the 5 subjects.

No. of ways in which student can fail in any 2 subjects out of the 5 subjects = 5C2

Case 3: Student fails in any 3 subjects out of the 5 subjects.

No. of ways in which student can fail in any 3 subjects out of the 5 subjects = 5C3

Case 4: Student fails in any 4 subjects out of the 5 subjects.

No. of ways in which student can fail in any 4 subjects out of the 5 subjects = 5C4

Case 5: Student fails in all 5 subjects.

No. of ways in which student can fail all 5 subjects. = 5C5

∴ Total number of ways in which a student can fail in an examination = 5C1 + 5C2 + 5C3 + 5C4 + 5C5 = 31

If 2nC3 : nC2 = 12 : 1, then find the value of n.

  1. 2
  2. 3
  3. 5
  4. 7

Answer (Detailed Solution Below)

Option 3 : 5

Combination Function and its properties Question 10 Detailed Solution

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Concept:

The number of all combinations of n distinct things taken r at a time (r ≤ n) is:

nCr=n!(nr)!r!

Calculation:

Since 2nC3 : nC2 = 12 : 1

2n!(2n3)! 3!:n!(n2)! 2!=12:1

2n(2n1)(2n2)3×2×1:n(n1)2×1=12:1

⇒ 2n - 1 = 9

⇒ n = 5

 If nC15 = nC8 ,then find the value of n.

  1. 24
  2. 23
  3. 21
  4. 20

Answer (Detailed Solution Below)

Option 2 : 23

Combination Function and its properties Question 11 Detailed Solution

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Concept:

  • nCr = nCn - r.
  • If nCx = nCy, then x + y = n.
  • nCr + nCr - 1 = n + 1Cr.


Calculation:

Given: nC15 = nC8

As we know, nCx = nCy, then x + y = n.

So, n = 15 + 8 = 23

Out of 7 consonants and 4 vowels, how many words can be formed such that it contains 3 consonants and 2 vowels ?

  1. 36000
  2. 55000
  3. 25200
  4. 75000

Answer (Detailed Solution Below)

Option 3 : 25200

Combination Function and its properties Question 12 Detailed Solution

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Concept:

The number of ways to select r things out of n given things wherein r ≤ n is given by: nCr=n!r!×(nr)!

Calculation:

Given: There are 7 consonants and 4 vowels

Here, we have to find how many word can be formed such that it has 3 consonants and 2 vowels.

No. of ways to select 2 vowels out of 4 vowels = 4C2

No. of ways to select 3 consonants out of 7 consonants = 7C3

∴ No. of words that can be formed which contains 3 consonants and 2 vowels = 4C2 × 7C3

As we know that, nCr=n!r!×(nr)!

⇒ 4C2 × 7C3  = 6 × 35 = 210

The no. of ways to arrange words containing 3 consonants and 2 vowels = 210 × 5! = 25200

Hence, option 3 is the correct answer

From a group of 10 men and 8 women, a team of 3 men or 2 women is to be made. How many teams can be made?

  1. 148
  2. 130
  3. 150
  4. 145

Answer (Detailed Solution Below)

Option 1 : 148

Combination Function and its properties Question 13 Detailed Solution

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Concept:

The number of ways to select r things out of n given things wherein r ≤ n is given by: nCr=n!r!×(nr)!

Calculation:

Given: The group has 10 men and 8 women.

Here, we need to form a team of 3 men or 2 women .

As we know that, the  number of ways to select r things out of n given things wherein r ≤ n is given by: nCr=n!r!×(nr)!

∴ No. of ways to form a team of 3 men =10C3=10!3!×(103)!

∴ No. of ways to form a team of 2 women =8C2=8!2!×(82)!

∴No. of ways to form a team of 3 men or 2 women =10C3+8C2=148

Out of 6 teachers and 8 students a committee of 11 is to be formed. In how many ways can this be done such that the committee has exactly 4 teachers ?

  1. 140
  2. 135
  3. 125
  4. 120

Answer (Detailed Solution Below)

Option 4 : 120

Combination Function and its properties Question 14 Detailed Solution

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Concept:

The number of ways to select r things out of n given things wherein r ≤ n is given by: nCr=n!r!×(nr)!

Calculation:

Given: There are 6 teachers and 8 students  out of which we need to form a committee of 11 members such that the committee has exactly 4 teachers in it.

If out 11 members of committee 4 are teachers then 7 members are students.

So, the committee compromises of 4 teachers and 7 students.

No. of ways in which 4 teachers can be selected from 6 teachers = 6C4

No. of ways in which 7 students can be selected from 8 students = 8C7

So, the number of ways in which the committee can be formed = 6C4×8C7

As we know that, nCr=n!r!×(nr)!

⇒ 6C4×8C7=120

Hence, the committee can be formed in 120 ways

Answer (Detailed Solution Below)

Option 4 : 4

Combination Function and its properties Question 15 Detailed Solution

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Concept:

Combinatorics:

  • nCr = n!r!(n  r)!.
  • n! = 1 × 2 × 3 × ... × n.
  • 0! = 1.


Calculation:

Given equation is: 3(6C3) = 10(xC2)

⇒ 3×[6!3!(6  3)!]=10×[x!2!(x  2)!]

⇒ 3×(6 × 5 × 43 × 2 × 1)=10×[x(x  1)2 × 1]

⇒ x(x - 1) = 12 = 4 × 3

x = 4.

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