Permutations and Combinations Test

Result

Permutations and Combinations Test - 1

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Weekly Quiz Competition

Question 1
2 / -0.83

In how many ways 4 boys and 3 girls can be seated in a row so that they are alternate.

A
144

B
288

C
12

D
256

Solution
Question 2
2 / -0.83

In how many ways 2 directors and 3 executives can be arranged for a meeting? If there are 6 chairs available two on one side and remaining four on the other side of the table and the two directors has to be together on one side and the executives on the other side.

A
24

B
48

C
720

D
120

Solution

3 executives in 4 chairs = 4P3 ways
Total no. of ways = 2! * 4P3
=>2*1*(4!/1!)
= 2*24
= 48
Question 3
2 / -0.83

A lady arranges a dinner party for 6 guests .The number of ways in which they may be selected from among 10 friends if 2 of the friends will not attend the party together is

A
164

B
140

C
112

D
none of these

Solution

Let the friends be A,B,C,D,E,F,G,H,I,J and assume A and B will not battend together
Case 1 : Both of them will not attend the party : Now we have to select the 6 guests from the remaing 8 members
Then no of ways is ⁸ C₆ = 28 ways.
Case 2 : Either of them are selected for the party :Now we have to select the 5 guests from the remaing 8 members and one from A and B
Then the no of ways = ² C₁ x ⁸ C₅ =112 ways.
Therefore total number of ways is 28+ 112 = 140 ways,
Question 4
2 / -0.83

What is the value of ⁿ p_o

A
zero

B
1

C
n

D
n!

Solution

nP0
= n!/(n-0)!
=>n!/n! = 1
Question 5
2 / -0.83

Find the value of n if ⁿ P₁ = 10

A
2

B
5

C
10

D
1

Solution

nPr = 10
nPr = n!/(n-r)!
10 = n!/(n-1)!
10 = [n!.(n-1)!]/(n-1)!
n! = 10
Question 6
2 / -0.83

Number of signals that can be made using 2 flags out of given 4 flags.

A
3

B
6

C
12

D
4

Solution

No. of ways of selecting two flags out of four = 4C2
So, total possible different signals generated = 4C2 ×2!
⟹6 ×2=12
Question 7
2 / -0.83

What is the value of ⁿ p_n

A
zero

B
n!

C
n

D
1

Solution

nPn
= n!/(n-n)!
= n!
Question 8
2 / -0.83

A
8

B
7

C
3

D
5

Solution

7! /5! ×3!
=7×6×5! /5! ×3×2×1
=7
Question 9
2 / -0.83

5 singers are to be given performance in a concert. How many ways can they be arranged in a order

A
60

B
24

C
48

D
120

Solution

5P5 ⇒5!/(5-5)!
= 5!/0!
= 5!/1 ⇒5!
= 120 ways
Question 10
2 / -0.83

What is the value of 0!

A
zero

B
1

C
not defined

D
2

Solution

The value of 0! = 1

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Re-Attempt Weekly Quiz Competition

Permutations and Combinations Test - 1

Result

Permutations and Combinations Test - 1

Score

-

Rank

-

SHARING IS CARING

Question 1

144

288

12

256

Solution

Question 2

24

48

720

120

Solution

Question 3

164

140

112

none of these

Solution

Question 4

zero

1

n

n!

Solution

Question 5

2

5

10

1

Solution

Question 6

3

6

12

4

Solution

Question 7

zero

n!

n

1

Solution

Question 8

8

7

3

5

Solution

Question 9

60

24

48

120

Solution

Question 10

zero

1

not defined

2

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

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