Online Test — Permutations and Combinations
10 Questions • 20 min • Chapter MCQ
20:00
Question 1 of 10
easy
The value of 11!/10! is:
10
11
110
1
Explanation: 11!/10! = 11.
Question 2 of 10
easy
Evaluate: 0! + 1! + 2! + 3!
10
9
6
12
Explanation: 0! = 1, 1! = 1, 2! = 2, 3! = 6. Sum = 1 + 1 + 2 + 6 = 10.
Question 3 of 10
medium
The Indian cricket board has to select 11 players out of a squad of 15. In how many ways can the team be selected?
1365
1001
3003
32760
Explanation: Selecting 11 out of 15 is 15C11. We know 15C11 = 15C4 = (15 × 14 × 13 × 12) / 24 = 1365.
Question 4 of 10
easy
A student can choose one of 5 mathematics books or one of 4 physics books. Total choices are:
9
20
10
1
Explanation: By addition principle, 5 + 4 = 9 choices.
Question 5 of 10
easy
The value of 6! is:
720
120
360
840
Explanation: 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720.
Question 6 of 10
easy
There are 3 different roads from City A to City B, and 4 different roads from City B to City C. In how many ways can a person travel from City A to City C via City B?
7
12
1
81
Explanation: By the Fundamental Principle of Multiplication, the total number of ways is 3 × 4 = 12.
Question 7 of 10
easy
The value of 12C11 is:
11
12
1
66
Explanation: 12C11 = 12C1 = 12.
Question 8 of 10
medium
The number of ways to arrange the letters of BOOK is:
24
12
6
8
Explanation: 4!/2! = 12.
Question 9 of 10
medium
The value of 7C5 is:
35
21
7
14
Explanation: 7C5 = 7C2 = 21.
Question 10 of 10
medium
Find the number of ways to select 4 cards from a standard deck of 52 cards such that all four are from different suits.
13⁴
52C4
13!
4!
Explanation: There are 4 suits, and we must select 1 card from each. From each suit, 1 card can be selected in 13C1 = 13 ways. Total ways = 13 × 13 × 13 × 13 = 13⁴.