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Class 11 Maths MCQ Questions of Permutations and Combinations with Answers?

Answer 1

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Practice MCQ Questions for class 11 Maths Chapter-Wise

1. There are 12 points in a plane out of which 5 are collinear. The number of triangles formed by the points as vertices is

(a) 185
(b) 210
(c) 220
(d) 175

2. The number of combination of n distinct objects taken r at a time be x is given by

(a) ^n/2C_r
(b) ^n/2C_r/2
(c) ⁿC_r/2
(d) ⁿC_r

3. How many factors are 2⁵ × 3⁶ × 5² are perfect squares

(a) 24
(b) 12
(c) 16
(d) 22

4. How many 3-letter words with or without meaning, can be formed out of the letters of the word, LOGARITHMS, if repetition of letters is not allowed

(a) 720
(b) 420
(c) none of these
(d) 5040

5. The number of ways in which 8 distinct toys can be distributed among 5 children is

(a) 5⁸         
(b) 8⁵
(c) ⁸P₅
(d) ⁵P₅  

6. The value of P(n, n – 1) is

(a) n
(b) 2n
(c) n!
(d) 2n!

7. In how many ways can 4 different balls be distributed among 5 different boxes when any box can have any number of balls?

(a) 5⁴ – 1
(b) 5⁴
(c) 4⁵ – 1
(d) 4⁵

8. The number of ways of painting the faces of a cube with six different colors is

(a) 1
(b) 6
(c) 6!
(d) None of these

9. 6 men and 4 women are to be seated in a row so that no two women sit together. The number of ways they can be seated is

(a) 604800
(b) 17280
(c) 120960
(d) 518400

10. The number of ways can the letters of the word ASSASSINATION be arranged so that all the S are together is

(a) 152100
(b) 1512
(c) 15120
(d) 151200

11. If repetition of the digits is allowed, then the number of even natural numbers having three digits is

(a) 250
(b) 350
(c) 450
(d) 550

12. How many ways are here to arrange the letters in the word GARDEN with the vowels in alphabetical order?

(a) 120
(b) 240
(c) 360
(d) 480

13. A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is

(a) 40
(b) 196
(c) 280
(d) 346

14. Four dice are rolled. The number of possible outcomes in which at least one dice show 2 is

(a) 1296
(b) 671
(c) 625
(d) 585

15. In how many ways in which 8 students can be sated in a line is

(a) 40230
(b) 40320
(c) 5040
(d) 50400

16. Given 12 points in a plane, no three of which are collinear. Then number of line segments can be determined, are:

(a) 76
(b) 66
(c) 60
(d) 80 

17. The total number of ways of selecting six coins out of 20 one rupee coins, 10 fifty paise coins and 7 twenty five paise coins is:

(a) 37
(b) 56
(c) 28
(d) 29

18. The number of numbers consisting of four different digits that can be formed with the digits 0, 1, 2, 3 is :

(a) 18
(b) 24
(c) 30
(d) 72

19. In a certain test, there are n questions. In this test 2n−i students gave wrong answers to at least i questions; where i=1,2.............n−1,n. If the total number of wrong answers given is 2047, then n is equal to

(a) 10
(b) 11
(c) 12
(d) 13

20. The sum of all five digit numbers that can be formed using the digits 1, 2, 3, 4, 5 when repetition of digits is not allowed, is

(a) 366000
(b) 660000
(c) 360000
(d) 3999960

Answer 2

1. Answer: (b) 210

Explanation: Total number of triangles that can be formed with 12 points (if none of them are collinear)
= ¹²C₃ (this is because we can select any three points and form the triangle if they are not collinear)
With collinear points, we cannot make any triangle (as they are in straight line). Here 5 points are collinear. Therefore we need to subtract ⁵C₃ triangles from the above count.
Hence, required number of triangles 

= ¹²C₃ – ⁵C₃ 

= 220 – 10 

= 210

2. Answer: (d) ⁿC_r

Explanation: The number of combination of n distinct objects taken r at a time be x is given by

ⁿC_r = n!/{(n – r)! × r!}

Let the number of combination of n distinct objects taken r at a time be x.

Now consider one of these n ways. There are e objects in this selection which can be arranged in r! ways.

So, each of the x combinations gives rise to r! permutations. So, x combinations will give rise to x×(r!).

Consequently, the number of permutations of n things, taken r at a time is x×(r!) and it is equal to ⁿP_r

So, x × (r!) = ⁿP_r

⇒ x×(r!) = n!/(n – r)!

⇒ x = n!/{(n – r)! × r!}

⇒ ⁿC_r = n!/{(n – r)! × r!}

3. Answer: (a) 24

Explanation: Any factors of 2⁵ × 3⁶ × 5² which is a perfect square will be of the form 2a × 3b × 5c where a can be 0 or 2 or 4, So there are 3 ways b can be 0 or 2 or 4 or 6, So there are 4 ways a can be 0 or 2, So there are 2 ways.

So, the required number of factors 

= 3 × 4 × 2 

= 24

4. Answer: (a) 720

Explanation: The word LOGARITHMS has 10 different letters.

Hence, the number of 3-letter words(with or without meaning) formed by using these letters

= ¹⁰P₃

= 10 ×9 ×8

= 720

5. Answer: (a) 5⁸  

Explanation: Total number of toys = 8

Total number of children = 5

Now, each toy can be distributed in 5 ways.

So, total number of ways

= 5 × 5 × 5 × 5 × 5 × 5 × 5 × 5

= 5⁸

6. Answer: (c) n!

Explanation: Given, P(n, n – 1)

= n!/{(n – (n – 1)}

= n!/(n – n + 1)}

= n!

So, P(n, n – 1) = n!

7. Answer: (b) 5⁴

Explanation: Here, both balls and boxes are different.

Now, 1st ball can be placed into any of the 5 boxes. 2nd ball can be placed into any of the 5 boxes. 3rd ball can be placed into any of the 5 boxes.4th ball can be placed into any of the 5 boxes.

So, the required number of ways

= 5 × 5 × 5 × 5 

= 5⁴

8. Answer: (a) 1

Explanation: Since the number of faces is same as the number of colors, therefore the number of ways of painting them is 1.

9. Answer: (a) 604800

Explanation: 6 men can be sit as × M × M × M × M × M × M ×

Now, there are 7 spaces and 4 women can be sit as ⁷P₄ = ⁷P₃ 

= 7!/3! 

= (7 × 6 × 5 × 4 × 3!)/3!

= 7 × 6 × 5 × 4 = 840

Now, total number of arrangement = 6! × 840

= 720 × 840

= 604800

10. Answer: (d) 151200

Explanation: Given word is : ASSASSINATION

Total number of words = 13

Number of A : 3
Number of S: 4
Number of I: 2
Number of N: 2
Number of T: 1
Number of O: 1 

Now all S are taken together. So it forms a single letter.

Now total number of words = 10

Now number of ways so that all S are together = 10!/(3!×2!×2!)

= (10×9×8×7×6×5×4×3!)/(3! × 2×2)

= (10×9×8×7×6×5×4)/(2×2)

= 10×9×8×7×6×5

= 151200

So total number of ways = 151200

Answer 3

11. Answer: (c) 450

Explanation: In a 3 digit number, 1st place can be filled in 5 different ways with (0, 2, 4, 6, 8)

10th place can be filled in 10 different ways.100th place can be filled in 9 different ways.

So, the total number of ways

= 5 × 10 × 9

= 450

12. Answer: (c) 360

Explanation: Given word is GARDEN.

Total number of ways in which all letters can be arranged in alphabetical order = 6!

There are 2 vowels in the word GARDEN A and E.

So, the total number of ways in which these two vowels can be arranged = 2!

Hence, required number of ways

= 6!/2!

= 720/2

= 360

13. Answer: (b) 196

Explanation: There are two cases

(i) When 4 is selected from the first 5 and rest 6 from remaining 8

Total arrangement = ⁵C₄ × ⁸C₆

= ⁵C₁ × ⁸C₂

= 5 × (8×7)/(2×1)

= 5 × 4 × 7

= 140

(ii) When all 5 is selected from the first 5 and rest 5 from remaining 8

Total arrangement = ⁵C₅ × ⁸C₅

= 1 × ⁸C₃

= (8×7×6)/(3×2×1)

= 8×7

= 56

Now, total number of choices available

= 140 + 56

= 196

14. Answer: (b) 671

Explanation: No. of ways in which any number appearing in one dice = 6

No. of ways in which 2 appear in one dice = 1

No. of ways in which 2 does not appear in one dice = 5

There are 4 dice.Getting 2 in at least one dice = Getting any number in all the 4 dice – Getting not 2 in any of the 4 dice.

= (6×6×6×6) – (5×5×5×5)

= 1296 – 625

= 671

15. Answer: (b) 40320

Explanation: The number of ways in which 8 students can be sated in a line = ⁸P₈

= 8!

= 40320

16. Answer: (b) 66

Explanation: n =¹²C₂ \(=\frac{12!}{2!(12!-2)!}\)

\(=\frac{12\times11\times10!}{2\times 10!}\)

= 66

17. Answer: (c) 28

Explanation: Since we know that the total number of selections of r things from n things where each thing can be repeated as many times as one can, is ^n+r−1C_r

Here r = 6 (∵ we have to select 6 coins) and n = 3 (∵ it is repeated 3 times), 

Required number 

= ³⁺⁶⁻¹C₆

= 28

18. Answer: (a) 18

Explanation: The number of numbers consisting of four different digits that can be formed with digits 0, 1, 2, 3 

= 4! - 3!

= 4. 3! - 3!

= 3!(4 - 1) = 3! . 3

⇒ 3.2.1.3 

⇒ 18

19. Answer: (b) 11

Explanation: The no. of students answering exactly i(1≤ i≤ n−1) questions wrongly is 2ⁿ⁻ⁱ −2ⁿ⁻ⁱ⁻¹. The no. of students answering all n questions wrongly is 2°. Thus, the total number of wrong answer is

= 1(2ⁿ⁻¹−2ⁿ⁻²)+2 (2ⁿ⁻²−2ⁿ⁻³)+…(n−1)(2¹−2°) +n (2°)

⇒ 2ⁿ⁻¹+2ⁿ⁻²+2ⁿ⁻³+.........+2+1

= 2ⁿ−1 

Thus 2ⁿ−1 = 2047

2ⁿ = 2048

= 2¹¹

∴n = 11

20. Answer: (d) 3999960

Explanation: Let the position of digits be as shown.

When 1 comes on Ist place then rest 4 place can be filled in ⁴P₄ way = 24 ways

i.e, 1 will come at first place, 24 times; Similarily for each digits at 2, 3, 4 & 5, at first place from left will come , 24 times:

Similarly for 2nd place, 3rd place, 4th place, and 5th place,

Hence sum of all five digit numbers.

24{10,000 (1 + 2 + 3 + 4 + 5 ) + 1000 (1 + 2 + 3 + 4 + 5) + 100 (1 + 2 + 3 + 4 + 5) + 10 (1 + 2 + 3 + 4 + 5) + (1 + 2 + 3 + 4 + 5)}

= 24 × 15{11111}

= 11111 × 360 

= 3999960

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