BITSAT Mathematics Permutations and Combinations MCQs

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MCQ for BITSAT Mathematics Permutations and Combinations

Review these 50 questions and answers for BITSAT Mathematics to improve your problem-solving skills for Permutations and Combinations.

Permutations and Combinations MCQ Questions BITSAT Mathematics with Answers

 

Question: How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even positions ?

  • a) 16
  • b) 36
  • c) 60
  • d) 180

Answer: 60

 

Question: Statement 1 : A five digit number divisible by 3 is to be formed using the digits 0, 1, 2, 3, 4 and 5 with repetition. The total number formed are 216. Statement 2 : If sum of digits of any number is divisible by 3 then the number must be divisible by 3.

  • a) Statement-1 is true, Statement-2 is true and is a correct explanation for Statement -1
  • b) Statement -1 is true, Statement -2 is true and is NOT a correct explanation for Statement - 1
  • c) Statement - 1 is true, Statement- 2 is false
  • d) Statement -1 is false, Statement -2 is true

Answer: Statement -1 is false, Statement -2 is true


Question: If

 BITSAT Mathematics Permutations 1

 then a – n is equal to

  • a) 0
  • b) 1
  • c) 2
  • d) None of these

Answer: 0 

Question: The number of values of r satisfying the equation BITSAT Mathematics Permutations 2is

  • a) 1
  • b) 2
  • c) 3
  • d) 4

Answer: 2

 

Question: All the words that can be formed using alphabets A, H, L, U and R are written as in a dictionary (no alphabet is repeated). Rank of the word RAHUL is

  • a) 71
  • b) 72
  • c) 73
  • d) 74

Answer: 74

 

Question: The total number of 4-digit numbers in which the digits are in descending order, is

  • a) 10C4 × 4!
  • b) 10C4
  • c)

    BITSAT Mathematics Permutations 3

  • d) None of these

Answer:   10C4

 

Question: The number of all three elements subsets of the set {a1, a2, a3 . . . an} which contain a3 is

  • a) nC3
  • b) n – 1C3
  • c) n – 1C2
  • d) None of these

Answer: n – 1C2

 

Question: In how many ways can a committee of 5 made out 6 men and 4 women containing atleast one woman?

  • a) 246
  • b) 222
  • c) 186
  • d) None of these

Answer: 246

 

Question: In how many ways can 5 boys and 5 girls be seated at a round table so that no two girls may be together ?

  • a) 4!
  • b) 5!
  • c) 4! + 5!
  • d) 4! × 5!

Answer: 4! × 5!

 

Question: A box contains two white balls, three black balls and four red balls. In how many ways can three balls be drawn from the box if at least one black ball is to be included in the draw?

  • a) 64
  • b) 129
  • c) 84
  • d) None of these

Answer: 64

 

Question: In how many ways can 5 prizes be distributed among 4 boys when every boy can take one or more prizes ?

  • a) 1024
  • b) 625
  • c) 120
  • d) 600

Answer: 1024

 

Question: The number of positive integral solution of abc= 30 is

  • a) 30
  • b) 27
  • c) 8
  • d) None of these

Answer: 27

 

Question: If 20Cr = 20Cr–10 then 18Cr is equal to

  • a) 4896
  • b) 816
  • c) 1632
  • d) None of these

Answer: 816

 

Question: In a polygon no three diagonals are concurrent. If the total number of points of intersection of diagonals interior to the polygon be 70 then the number of diagonals of the polygon is

  • a) 20
  • b) 28
  • c) 8
  • d) None of these

Answer: 20

 

Question: With 17 consonants and 5 vowels the number of words of four letters that can be formed having two different vowels in the middle and one consonant, repeated or different at each end is

  • a) 5780
  • b) 2890
  • c) 5440
  • d) 2720

Answer: 5780

 

Question: The letters of the word TOUGH are written in all possible orders and these words are written out as in a dictionary, then the rank of the word TOUGH is

  • a) 120
  • b) 88
  • c) 89
  • d) 90

Answer: 89

 

Question: If 2n +1Pn–1 : 2n – 1Pn = 3 : 5 then the value of n is equal to

  • a) 4
  • b) 3
  • c) 2
  • d) 1

Answer: 4

 

Question: 3 integers are chosen at random from the set of first 20 natural numbers. The chance that their product is a multiple of 3, is –

  • a) 194/285
  • b) 1/57
  • c) 13/19
  • d) 3/4

Answer: 194/285

 

Question: If the total number of m elements subsets of the set A = {a1, a2, a3, ...., an} is l times the number of 3 elements subsets containing a4, then n is

  • a) (m – 1) λ
  • b) mλ
  • c) (m + 1) λ
  • d) 0

Answer: 

 

Question: The number of arrangements of the letters of the word BANANA is which the two ‘N’s do not appear adjacently is

  • a) 40
  • b) 60
  • c) 80
  • d) 100

Answer: 40

 

Permutations and Combinations Objective Questions & Solutions for BITSAT Mathematics

Practice MCQs: Permutations and Combinations (BITSAT)

Utilize these MCQs for Permutations and Combinations to test your mastery of the chapter efficiently. Formatted under recent BITSAT guidelines for BITSAT Mathematics, these multiple-choice exercises support steady learning. Working through these objective questions daily leads to better academic performance.

Important Objective Questions & Solutions for Permutations and Combinations

Compiled directly from the official NCERT book for BITSAT, these Mathematics MCQs focus on high-yield exam areas frequently tested in evaluations. Once finished, cross-reference your answers with our given solutions. To deepen your understanding of Permutations and Combinations, read through our professional NCERT solutions for BITSAT Mathematics.

Interactive MCQ Tests for BITSAT Mathematics

To prepare for your exams you should also take the BITSAT Mathematics MCQ test for this chapter on our website. This will help you improve your speed and accuracy and it is also free for you. Regular revision of these Mathematics topics will make you an expert in all important chapters of your course.

FAQs

Where can I access latest BITSAT Mathematics Permutations and Combinations MCQs?

You can get most exhaustive BITSAT Mathematics Permutations and Combinations MCQs for free on StudiesToday.com. These MCQs for BITSAT Mathematics are updated for the 2026-27 academic session as per BITSAT examination standards.

Are Assertion-Reasoning and Case-Study MCQs included in the Mathematics BITSAT material?

Yes, our BITSAT Mathematics Permutations and Combinations MCQs include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the BITSAT paper is now competency-based.

How do practicing Mathematics MCQs help in scoring full marks in BITSAT exams?

By solving our BITSAT Mathematics Permutations and Combinations MCQs, BITSAT students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Mathematics.

Do you provide answers and explanations for BITSAT Mathematics Permutations and Combinations MCQs?

Yes, Mathematics MCQs for BITSAT have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused BITSAT exams.

Can I practice these Mathematics BITSAT MCQs online?

Yes, you can also access online interactive tests for BITSAT Mathematics Permutations and Combinations MCQs on StudiesToday.com as they provide instant answers and score to help you track your progress in Mathematics.