Permutations and Combinations Class 11 MCQ Online Test With Answers Questions

Answer: (b) 210
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

Answer: (d) ⁿC_r
Hint:
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!}

Answer: (b) 671
Hint:
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

Answer: (c) 450
Hint:
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

Answer: (a) 5⁸
Hint:
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

Answer: (c) n!
Hint:
Given,
Given, P(n, n – 1)
= n!/{(n – (n – 1)}
= n!/(n – n + 1)}
= n!
So, P(n, n – 1) = n!

Answer: (b) 5⁴
Hint:
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

Answer: (a) 1
Hint:
Since the number of faces is same as the number of colors,
therefore the number of ways of painting them is 1

Answer: (c) 1008
Hint:
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+1C_r × x^r
= {(1- x¹¹ – x⁶ + x¹⁷)×{(1- x¹⁴)} × ∑^3+r+1C_r × x^r
= {(1- x¹¹ – x⁶ + x¹⁷ – x¹⁴ + x²⁵ + x²⁰ – x³¹)} × ∑^2+rC_r × x^r
= 1 × ∑^2+rC_r × x^r – x¹¹ × ∑^2+rC_r × x^r – x⁶ × ∑^2+rC_r × x^r + x¹⁷ × ∑^2+rC_r × x^r – x¹⁴ × ∑^2+rC_r × x^r + x²⁵ × ∑^2+rC_r × x^r + x²⁰ × ∑^2+rC_r × x^r – x³¹ × ∑^2+rC_r × x^r
= ∑^2+rC_r × x^r – ∑^2+rC_r × x^r+11 – ∑^2+rC_r × x^r+6 + ∑^2+rC_r × x^r+17 – ∑^2+rC_r × x^r+14 + ∑^2+rC_r × x^r+25 + ∑^2+rC_r × x^r+20 – ∑^2+rC^r × x^r+25
Now we have to find co-efficeient 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

Answer: (a) 604800
Hint:
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

Answer: (d) 151200
Hint:
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: (c) 450
Hint:
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

Answer: (b) 7
Hint:
The number of triangles that can be formed using the vertices of a regular polygon = ⁿC₃
Given, T_n+1 – T_n = 21
⇒ ⁿ⁺¹C₃ – ⁿC₃ = 21
⇒ ⁿC₂ + ⁿC₃ – ⁿC₃ = 21 {since ⁿ⁺¹C_r = ⁿC_r-1 + ⁿC_r}
⇒ ⁿC₂ = 21
⇒ n(n – 1)/2 = 21
⇒ n(n – 1) = 21×2
⇒ n² – n = 42
⇒ n² – n – 42 = 0
⇒ (n – 7)×(n + 6) = 0
⇒ n = 7, -6
Since n can not be negative,
So, n = 7

Answer: (c) 360
Hint:
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

Answer: (a) 24
Hint:
Any factors of 2⁵ × 3⁶ × 5² which is a perfect square will be of the form 2^a × 3^b × 5^c
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

Answer: (b) 196
Hint:
There are two cases
1. 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
2. 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

Answer: (b) 671
Hint:
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

Answer: (b) 40320
Hint:
The number of ways in which 8 students can be sated in a line = ⁸P₈
= 8!
= 40320

Answer: (d) 204
Hint:
A chess board contains 9 lines horizontal and 9 lines perpendicular to them.
To obtain a square, we select 2 lines from each set lying at equal distance and this equal
distance may be 1, 2, 3, …… 8 units, which will be the length of the corresponding square.
Now, two lines from either set lying at 1 unit distance can be selected in ⁸C₁ = 8 ways.
Hence, the number of squares with 1 unit side = 8²
Similarly, the number of squares with 2, 3, ….. 8 unit side will be 7², 6², …… 1²
Hence, total number of square = 8² + 7² + ……+ 1² = 204

Answer: (a) 720
Hint:
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