Permutations and Combinations (BITSAT Mathematics): Questions 7  13 of 13
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Question number: 7
» Permutations and Combinations » Simple Applications
Question
is equal to
Choices
Choice (4)  Response  

a. 


b. 


c. 


d.  All of the above 

Question number: 8
» Permutations and Combinations » Permutation as an Arrangement
Question
If , then the value of is
Choices
Choice (4)  Response  

a.  2 or 6 

b.  6 or 7 

c.  7 or 11 

d.  2 or 11 

Question number: 9
» Permutations and Combinations » Combination as Selection
Question
A box contains two white balls, three black balls and four red balls. The number of ways in which three balls can be drawn from the box so that one of the balls is black is –
Choices
Choice (4)  Response  

a.  20 

b.  64 

c.  84 

d.  74 

Question number: 10
» Permutations and Combinations » Simple Applications
Question
The value of expression
is equal to –
Choices
Choice (4)  Response  

a. 


b. 


c. 


d. 


Question number: 11
» Permutations and Combinations » Fundamental Principle of Counting
Question
There are copies of each different books in library. The number of ways in which one or more than one book can be selected as –
Choices
Choice (4)  Response  

a. 


b. 


c. 


d. 


Question number: 12
» Permutations and Combinations » Fundamental Principle of Counting
Question
The number of ways in which one or more balls can be selected out of 10 white, 9 green and 7 blue balls, is –
Choices
Choice (4)  Response  

a.  881 

b.  892 

c.  879 

d.  891 

Question number: 13
» Permutations and Combinations » Simple Applications
Question
The number of all 3 elements subsets of which contains is –
Choices
Choice (4)  Response  

a. 


b. 


c. 


d.  All of the above 
