CBSE Class 11 Mathematics Chapter 7 Permutations and Combinations Multiple Choice Questions with Answers. MCQ Class 11 Mathematics Permutations and Combinations with Answers was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 11 Mathematics Permutations and Combinations MCQs with Answers to know their preparation level.
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MCQ Class 11 Mathematics Permutations and Combinations with Answers - Set - 1
Question 1:
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
Correct Answer – (A)
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
Question 2 :
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
Correct Answer – (A)
Question 3 :
The value of P(n, n – 1) is
(a) n
(b) 2n
(c) n!
(d) 2n!
Correct Answer – (C)
Given,
Given, P(n, n – 1)
= n!/{(n – (n – 1)}
= n!/(n – n + 1)}
= n!
So, P(n, n – 1) = n!
Question 4 :
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
Correct Answer – (C)
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
Question 5 :
The number of combination of n distinct objects taken r at a time be x is given by
(a) n/2Cr
(b) n/2Cr/2
(c) nCr/2
(d) nCr
Correct Answer – (D)
The number of combination of n distinct objects taken r at a time be x is given by
nCr = 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 nPr
So, x×(r!) = nPr
⇒ x×(r!) = n!/(n – r)!
⇒ x = n!/{(n – r)! × r!}
⇒ nCr = n!/{(n – r)! × r!}
MCQ Class 11 Mathematics Permutations and Combinations with Answers
Question 6 :
Out of 5 apples, 10 mangoes and 13 oranges, any 15 fruits are to be distributed among 2 persons. Then the total number of ways of distribution is
(a) 1800
(b) 1080
(c) 1008
(d) 8001
Correct Answer – (C)
Given there are 5 apples, 10 mangoes and 13 oranges.
Let x1 is for apple, x2 is for mango and x3 is for orange.
Now, first we have to select total 15 fruits out of them.
x1 + x2 + x3 = 15 (where 0 ⇐ x1 ⇐ 5, 0 ⇐ x2 ⇐ 10, 0 ⇐ x3 ⇐ 13)
= (x0 + x1 + x2 +………+ x5)×(x0 + x1 + x2 +………+ x110)×(x0 + x1 + x2 +………+ x13)
= {(1- x6)/(1 – x)}×{(1- x11)/(1 – x)}×{(1- x14)/(1 – x)}
= {(1- x6)×(1- x11)×{(1- x14)}/(1 – x)³
= {(1- x6)×(1- x11)×{(1- x14)} × ∑3+r+1Cr × xr
= {(1- x11 – x6 + x17)×{(1- x14)} × ∑3+r+1Cr × xr
= {(1- x11 – x6 + x17 – x14 + x25 + x20 – x31)} × ∑2+rCr × xr
= 1 × ∑2+rCr × xr – x11 × ∑2+rCr × xr – x6 × ∑2+rCr × xr + x17 × ∑2+rCr × xr – x14 × ∑2+rCr × xr + x25 × ∑2+rCr × xr + x20 × ∑2+rCr × xr – x31 × ∑2+rCr × xr
= ∑2+rCr × xr – ∑2+rCr × xr+11 – ∑2+rCr × xr+6 + ∑2+rCr × xr+17 – ∑2+rCr × xr+14 + ∑2+rCr × xr+25 + ∑2+rCr × xr+20 – ∑2+rCr × xr+25
Now we have to find co-efficient of x15
= 2+15C15 – 2+4C4 – 2+9C9 – 2+1C1 (rest all terms have greater than x15, so its coefficients are 0)
= 17C15 – 6C4 – 11C9 – 3C1
= 17C2 – 6C2 – 11C2 – 3C1
= {(17×16)/2} – {(6×5)/2} – {(11×10)/2} – 3
= (17×8) – (3×5) – (11×5) – 3
= 136 – 15 – 55 – 3
= 136 – 73
= 63
Again we have to distribute 15 fruits between 2 persons.
So x1 + x2 = 15
= 2-1+15C15
= 16C15
= 16C1
= 16
Now total number of ways of distribution = 16 × 63 = 1008
Question 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
Correct Answer – (B)
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
Question 8 :
The number of ways in which 8 distinct toys can be distributed among 5 children is
(a) 58
(b) 85
(c) 8P5
(d) 5P5
Correct Answer – (A)
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
Question 9 :
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
Correct Answer – (B)
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
Question 10 :
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
Correct Answer – (B)
Hint:
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
- NCERT Solutions Class 11 Mathematics Permutations and Combinations with Answers : Exercise 7.1
- NCERT Solutions Class 11 Mathematics Permutations and Combinations with Answers : Exercise 7.2
- NCERT Solutions Class 11 Mathematics Permutations and Combinations with Answers : Exercise 7.3
- NCERT Solutions Class 11 Mathematics Permutations and Combinations with Answers : Exercise 7.4
- NCERT Solutions Class 11 Mathematics Permutations and Combinations with Answers : Exercise 7 Misc
- NCERT Solutions Class 11 Mathematics Textbook download
