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Test: Combinations - JEE MCQ


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20 Questions MCQ Test Mathematics (Maths) for JEE Main & Advanced - Test: Combinations

Test: Combinations for JEE 2024 is part of Mathematics (Maths) for JEE Main & Advanced preparation. The Test: Combinations questions and answers have been prepared according to the JEE exam syllabus.The Test: Combinations MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Combinations below.
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Test: Combinations - Question 1

C(n,12) = C(n,8) N = ?

Detailed Solution for Test: Combinations - Question 1

nC12 = nC8
n/ 12 (n - 12) = n/8 (n - 8)
12 = n - 8
n = 12 + 8 = 20
n = 20.

Test: Combinations - Question 2

In a college there are 20 professors including the principal and the vice principal. A committee of 5 is to be formed. In how many ways it can be formed so that neither the principal nor the vice principal is included?

Detailed Solution for Test: Combinations - Question 2

Out of 20 professors, excluding the principal and vice-principal, 18 professors are left.
No. of ways of selecting 5 professors out of 18 = 18C5

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Test: Combinations - Question 3

What is the number of diagonals that can be drawn by joining the vertices of a hexagon?

Detailed Solution for Test: Combinations - Question 3

No. of diagonals in a n-sided polygon = n *(n – 3)/2      
No. of diagonals in a hexagon = 6 *(6 – 3)/2 = 9

Test: Combinations - Question 4

Nidhi has 6 friends. In how many ways can she invite one or more of them to a party at her home?

Detailed Solution for Test: Combinations - Question 4

She has 6 friends and he wants to invite one or more. That is the same as saying he wants to invite at least 1 of his friends.
 
So, the number of ways he could do this is:
Invite only one friend
Invite any two friends
Invite any three friends
Invite any four friends
Invite any five friends
Invite all six friends
This can be thought of in terms of combinations. Inviting  r  friends out of  n  is same as choosing  r  friends out of  n . So, we can write the possibilities as:
6C1 + 6C2 + 6C3 + 6C4 + 6C5 + 6C6 
= 6 + 15 + 20 + 15 + 6 + 1 
= 63

Test: Combinations - Question 5

If C(n,4) = 495. Find n.

Detailed Solution for Test: Combinations - Question 5

nC4 = 495
n!/r!(n-r)! = 495
n!/4!(n-4)! = 495
n!/24(n-4)! = 495
n!/(n-4)! = 495 * 24
n!/(n-4)! = 11850
[n(n-1)(n-2)(n-3)(n-4)!]/(n-4)! = 11850
[n(n-1)(n-2)(n-3)] = 12*11*10*9
n = 12

Test: Combinations - Question 6

If nC8 = nC2, then n is

Detailed Solution for Test: Combinations - Question 6

Given nC8 = nC2
if nCr = nCp
Then either r=p or r=n−p
Thus, nC8 = nC2
8=n−2
10=n
∴ n=10

Test: Combinations - Question 7

8Cr = 8Cp. So

Detailed Solution for Test: Combinations - Question 7

8Cr = 8Cp = 8C8-p     
So, either r = p or r = 8 – p
i.e. r = p or r + p = 8

Test: Combinations - Question 8

What is the number of ways of choosing 6 cards from a pack of 52 playing cards?

Detailed Solution for Test: Combinations - Question 8

Test: Combinations - Question 9

If  = c. Find c.

Detailed Solution for Test: Combinations - Question 9

18P4/18C4
⇒ [18!/14!]/[18!/(4!*14!)]
⇒ 4!
⇒ 24

Test: Combinations - Question 10

In how many ways can a cricket team of 11 players selected out of 16 players if two particular players are to be included and one particular player is to be rejected?

Detailed Solution for Test: Combinations - Question 10

11 players can be selected out of 16 in 16C11 ways = 16!/(11! 5!) = 4368 ways. Now, If two particular players is to be included and one particular player is to be rejected, then we have to select 9 more players out of 13 in 13C9 ways.
= 13!/(9! 4!) = 715 ways.

Test: Combinations - Question 11

How many chords can be drawn through 10 points on a circle?

Detailed Solution for Test: Combinations - Question 11

For the 1st point chords can be drawn with joining 9 other points.
For the 2nd point distinct chords can be drawn with joining 8 other points.
 For the 9th point distinct chords can be drawn with joining 1 other point. 
So, total no. of points = 9 + 8 + 7 + . . . . . . + 1 = 45 

Test: Combinations - Question 12

In a room there are 2 green chairs, 3 yellow chairs and 4 blue chairs. In how many ways can Raj choose 3 chairs so that at least one yellow chair is included? 

Detailed Solution for Test: Combinations - Question 12

At least one yellow chair means “total way no yellow chair”
Total ways to select 3 chair form the total (2+3+4) = 9C3

Now, we dont want even one yellow chair.
So we should select 3 chairs from 6 chairs (2 green & 4 blue) = 6C3

Hence, ways to select at least one yellow chair = 84 - 20 = 64 ways.

Test: Combinations - Question 13

In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 courses are compulsory for every student.

Detailed Solution for Test: Combinations - Question 13

Test: Combinations - Question 14

C(20,r) = C(20, r + 2) What is the value of C(r,6) ?

Detailed Solution for Test: Combinations - Question 14

20Cr = 20Cr+2
nCr = nCn-r
=> r + (n−r) = n
⇒ r + r + 2 = 20
⇒ 2r = 18
⇒ r = 9
rC6 = 9C6
⇒ 9!/(6!*3!)
= 84

Test: Combinations - Question 15

A group consists of 4 girls and 7 boys. in how many ways a team of 5 members be selected consisting of 2 girls and 3 boys.

Detailed Solution for Test: Combinations - Question 15

Total number of girls in the group = 4
Total number of boys in the group = 7
Probability of selecting atleast three boy and two girl = 4C2 × 7C3
= 210

Test: Combinations - Question 16

In how many ways can an examinee choose 5 questions out of 8 in a paper of Physics?

Detailed Solution for Test: Combinations - Question 16

8C5 
= 8!/5!*3!
= 8*7*6*5!/5!*3!
= 56

Test: Combinations - Question 17

How many triangles can be formed by joining four points on a circle?

Detailed Solution for Test: Combinations - Question 17

Triangle formed = mC3
Therefore, 4 points on circle, m = 4
4C3 = 4C1
⇒ 4

Test: Combinations - Question 18

In how many ways can one select a cricket team of eleven from 13 players in which only 5 players can bowl. If each cricket team of 11 must include at least 4 bowlers.

Detailed Solution for Test: Combinations - Question 18

Test: Combinations - Question 19

A committee consists of 5 students 3 girls and 2 boys. If the team is to be formed out of 7 boys and 5 girls, then how many ways this selection can be made?

Detailed Solution for Test: Combinations - Question 19

out of 7 persons , we have to select exactly 3 girls . hence, it's clear that remaining 4 persons will be boys .
 
Now, number of ways taken 4 boys from 9 boys = 9C4
and number of ways taken 3 girls from 4 girls = 4C3
 
hence, by fundamental principle of counting ,
total number of ways for making the committee = 9C4 × 4C3
= 9!/5!.4! × 4!/3!
= 9×8×7×6/3×2×1
= 9×8×7
= 504 ways

Test: Combinations - Question 20

Find the number of diagonals of an n-sided polygon.

Detailed Solution for Test: Combinations - Question 20

For a convex n-sided polygon, there are n vertices, and from each vertex you can draw n-3 diagonals, so the total number of diagonals that can be drawn is n(n-3).The number of diagonals in a polygon = n(n-3)/2, where n is the number of polygon sides.

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