Free practice › Class 11 › Mathematics › Chapter 6
Permutations and Combinations — JEE MCQs with answers
Topics in this chapter
- 6.1 Introduction
- 6.2 Fundamental Principle of Counting
- 6.3 Permutations
- 6.4 Combinations
Free practice questions
According to $\S$6.4, what is the value of $^4\mathrm{C}_2$, the number of combinations of 4 different objects taken 2 at a time?
- A. 16
- B. 6
- C. 24
- D. 12
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Answer: B — NCERT lists the combinations AB, AC, AD, BC, BD, CD and concludes that there are as many as 6 of them, so option B is correct. Option D gives 12, which is $^4\mathrm{P}_2 = 4\times 3$, the ordered count; it is twice the answer because each combination is counted in both orders. Option A gives 16, which is $4\times 4$ and permits an object to be taken twice. Option C gives 24, which is $4!$, the number of arrangements of all four objects rather than the number of selections of two.
NCERT Class 11 Mathematics · Ch 6 · §6.4 ("included BA, CA, DA, CB, DB and DC in this list. There are as many as 6 combinations")
Beside the line $n! = n(n-1)(n-2)!$, NCERT prints a condition on $n$. What is it?
- A. $n \ge 3$
- B. $n \ge 0$
- C. $n \ge 1$
- D. $n \ge 2$
Show answer
Answer: D — Page 105 prints $n! = n(n-1)(n-2)!$ followed by the bracketed proviso $[$provided $(n \ge 2)]$, because the expression uses $(n-2)!$ and that needs $n-2 \ge 0$. Hence option D is correct. Option A ($n \ge 3$) is what reading across to the next printed line produces: NCERT attaches $[$provided $(n \ge 3)]$ to $n! = n(n-1)(n-2)(n-3)!$, which uses $(n-3)!$. The line with $(n-2)!$ needs only $n - 2 \ge 0$, i.e. $n \ge 2$. Option B ($n \ge 0$) is what allowing every whole number produces. At $n = 0$ and $n = 1$ the right side needs $(-2)!$ and $(-1)!$ respectively, neither of which is defined. Option C ($n \ge 1$) is what forgetting that $(n-2)!$ must itself make sense produces. At $n = 1$ the right side would need $(1-2)! = (-1)!$, which is not defined, so $n = 1$ is excluded.
NCERT Class 11 Mathematics · Ch 6 · §6.3.2 ("provided (n ≥ 2)")
Section 6.1 poses the number-lock question but prints no number for it, so the arithmetic below is worked out here rather than taken from NCERT. A student answers the question with 729. Which slip produces exactly that number?
- A. filling the three places from all 10 digits with no repetition: 10 x 9 x 8 = 720
- B. counting selections of 3 digits from the 9 remaining ones with no regard to order: 84
- C. counting arrangements of all 4 places from all 10 digits with no repetition: 10 x 9 x 8 x 7 = 5040
- D. allowing the three unknown digits to repeat, so that each of the three places may take any of the 9 remaining digits: 9 x 9 x 9 = 729
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Answer: D — Cubing 9 gives 729 exactly, and that is what allowing repetition among the 9 available digits does. The other slips produce different numbers and so cannot account for 729: filling the three places from all 10 digits gives 720, counting unordered selections of 3 from 9 gives 84, and arranging all four places out of 10 digits gives 5040. Section 6.1 forbids repetition, so 729 is not the count the lock question asks for in any case. None of this arithmetic is NCERT's: Section 6.1 poses the lock question and leaves it for the counting techniques that follow, so neither 729 nor the true count 504 is printed there. The correct option is D.
NCERT Class 11 Mathematics · Ch 6 · §6.1 ("particular sequence with no repetition.")
A student answers Example 1 by writing 4 × 4 × 4 × 4 = 256. What has the student overlooked, and what is the correct count?
- A. That only 3 places have to be filled, so the correct count is 64
- B. That the four counts must be added rather than multiplied, so the correct count is 16
- C. Nothing has been overlooked; 256 is the answer to Example 1
- D. That the repetition of the letters is not allowed, so the counts fall 4, 3, 2, 1 and the correct count is 24
Show answer
Answer: D — Option D is correct. Example 1 states that the repetition of the letters is not allowed, so each letter is used up as it is placed and the stage counts fall to 4, 3, 2, 1, giving 4 × 3 × 2 × 1 = 24. The word has 4 letters and 4 places, so it is not a case of only 3 places being needed, and 4 × 4 × 4 = 64 is not the answer either. Adding the stage counts would give 4 + 4 + 4 + 4 = 16, but the principle multiplies them. And 256 is not the answer to Example 1: it is the answer to the Note that follows, where repetition is expressly allowed.
NCERT Class 11 Mathematics · Ch 6 · §6.2 ("by the 4 letters, keeping in mind that the repetition is not allowed")
NCERT leaves the four-different-suits answer of Example 19 in the form $13^{4}$. Evaluated, this equals
- A. 2197
- B. 371293
- C. 28561
- D. 52
Show answer
Answer: C — Option C is correct. $13^{2} = 169$, $13^{3} = 169 \times 13 = 2197$ and $13^{4} = 2197 \times 13 = 28561$. The value 2197 is $13^{3}$, which would use only three suits. The value 371293 is $13^{5}$, one suit too many. The value 52 comes from multiplying $13 \times 4$ instead of raising 13 to the fourth power; it is the size of the whole pack, not the count required.
NCERT Class 11 Mathematics · Ch 6 · §6.4 ("There are13 cards in each suit.")
The letters of the word ROOT can be rearranged in how many distinct ways?
- A. 4
- B. 12
- C. 24
- D. 6
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Answer: B — Option B is correct. ROOT has 4 letters and the letter O appears twice, so the count is $4!/2!$, which the book writes as $3 \times 4 = 12$ and then confirms by listing all 12 words. Failing to divide gives $4! = 24$, which counts every word twice, since RO$_1$O$_2$T and RO$_2$O$_1$T are the same word ROOT. Fusing the two Os into a single block gives $3! = 6$, but that counts only the words whose two Os stand together, and ROTO, which appears in the printed list, has its Os apart. Treating O as occurring three times gives $4!/3! = 4$; the letters are R, O, O, T, so O occurs twice.
NCERT Class 11 Mathematics · Ch 6 · §6.3 ("the letters of the word are not all different")
A canteen offers 4 kinds of sandwich and 3 kinds of juice, and a meal is one sandwich together with one juice. Applying the fundamental principle of counting exactly as NCERT applies it to a pant and a shirt, how many different meals are possible?
- A. 12
- B. 24
- C. 16
- D. 42
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Answer: A — There are 4 ways in which a sandwich can be chosen, following which a juice can be chosen in 3 ways, so the meals number 4 × 3 = 12. Option A (12) is correct. The value 24 is what counting each meal twice gives, once from the sandwiches' side as 4 × 3 and again from the juices' side as 3 × 4; a meal is a single sandwich taken with a single juice and is not two meals. 16 uses the sandwich count for both stages, 4 × 4; the juice stage offers 3 choices, not 4. The value 42 comes of pooling the 4 sandwiches and the 3 juices into 7 items and picking 2 of them in order, 7 × 6 = 42; that count admits picking one sandwich and then another sandwich, which is not a meal at all.
NCERT Class 11 Mathematics · Ch 6 · §6.2 ("For every choice of a pant, there are 2 choices of a shirt")
The epigraph printed at the head of Chapter 6 reads: "Every body of discovery is mathematical in form because there is no other guidance we can have". To whom does NCERT attribute it?
- A. Jacob Bernoulli
- B. Blaise Pascal
- C. DARWIN
- D. Leonhard Euler
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Answer: C — The line is printed with the attribution DARWIN. Jacob Bernoulli is the mathematician whose portrait appears on the same page, but the quotation is not credited to him. Blaise Pascal and Leonhard Euler do not appear on this page at all. The correct option is C.
NCERT Class 11 Mathematics · Ch 6 · §6.1 ("other guidance we can have - DARWIN")
Remark 3 concludes that the formula $^{n}\mathrm{C}_{r}$ $= \dfrac{n!}{r!\,(n-r)!}$ is applicable for
- A. $r = n$ only
- B. $r > n$ also
- C. $n = 0$ only
- D. $r = 0$ also
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Answer: D — The remark's own words are “is applicable for $r = 0$ also”, so option D is correct. Nothing in the remark extends the formula to $r > n$; the range it then prints stops at $r = n$. The remark is about the lower end of the range of $r$, not about a single value of $n$, so restricting it to $n = 0$ is wrong. And $r = n$ was already handled in Remark 1, so it is not what Remark 3 is adding.
NCERT Class 11 Mathematics · Ch 6 · §6.4 ("is applicable for r = 0 also.")
In the worked solution for ALLAHABAD, $\dfrac{9!}{4!\,2!}$ is first rewritten as
- A. $\dfrac{6 \times 7 \times 8 \times 9}{2}$
- B. $\dfrac{5 \times 6 \times 7 \times 8 \times 9}{24}$
- C. $\dfrac{5 \times 6 \times 7 \times 8 \times 9}{2}$
- D. $\dfrac{5 \times 6 \times 7 \times 8 \times 9}{4}$
Show answer
Answer: C — Option C is correct. Cancelling $4!$ against the tail of $9!$ leaves $5 \times 6 \times 7 \times 8 \times 9 = 15120$, and the divisor still to be applied is $2! = 2$, giving $15120/2 = 7560$, the printed value. Dividing that product by 4 instead of by $2!$ gives 3780. Cancelling $5!$ instead of $4!$ leaves $6 \times 7 \times 8 \times 9 = 3024$, and halving gives 1512. Dividing the product by 24 gives 630. None of the three equals 7560.
NCERT Class 11 Mathematics · Ch 6 · §6.3 ("Example 9 Find the number of permutations of the letters of the word ALLAHABAD.")
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