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## Sunday, 21 February 2021

### CBSE Class 11 Maths - MCQ and Online Tests - Unit 7 - Permutations and Combinations

#### CBSE Class 11 Maths  – MCQ and Online Tests – Unit 7 – Permutations and Combinations

Every year CBSE schools conducts Annual Assessment exams for 6,7,8,9,11th standards. These exams are very competitive to all the students. So our website provides online tests for all the 6,7,8,9,11th standard’s subjects. These tests are also very effective and useful for those who preparing for any competitive exams like Olympiad etc. It can boost their preparation level and confidence level by attempting these chapter wise online tests.

These online tests are based on latest CBSE syllabus. While attempting these our students can identify the weak lessons and continuously practice those lessons for attaining high marks. It also helps to revise the NCERT textbooks thoroughly.

#### CBSE Class 11 Maths  – MCQ and Online Tests – Unit 7 – Permutations and Combinations

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

Hint:
6 men can be sit as
в M в M в M в M в M в M в
Now, there are 7 spaces and 4 women can be sit as 7P4 = 7P3 = 7!/3! = (7 в 6 в 5 в 4 в 3!)/3!
= 7 в 6 в 5 в 4 = 840
Now, total number of arrangement = 6! в 840
= 720 в 840
= 604800

Question 2.
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

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!в2!в2!)
= (10в9в8в7в6в5в4в3!)/(3! в 2в2)
= (10в9в8в7в6в5в4)/(2в2)
= 10в9в8в7в6в5
= 151200
So total number of ways = 151200

Question 3.
The number of ways in which 8 distinct toys can be distributed among 5 children is
(a) 58
(b) 85
(c) 8P5
(d) 5P5

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 в 5 в 5 в 5 в 5 в 5 в 5 в 5
= 58

Question 4.
The value of P(n, n – 1) is
(a) n
(b) 2n
(c) n!
(d) 2n!

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

Question 5.
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

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 в 5 в 5 в 5 = 54

Question 6.
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

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

Question 7.
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

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)в(x0 + x1 + x2 +………+ x110)в(x0 + x1 + x2 +………+ x13)
= {(1- x6)/(1 – x)}в{(1- x11)/(1 – x)}в{(1- x14)/(1 – x)}
= {(1- x6)в(1- x11)в{(1- x14)}/(1 – x)Ё
= {(1- x6)в(1- x11)в{(1- x14)} в ?3+r+1Cr в xr
= {(1- x11 – x6 + x17)в{(1- x14)} в ?3+r+1Cr в xr
= {(1- x11 – x6 + x17 – x14 + x25 + x20 – x31)} в ?2+rCr в xr
= 1 в ?2+rCr в xr – x11 в ?2+rCr в xr – x6 в ?2+rCr в xr + x17 в ?2+rCr в xr – x14 в ?2+rCr в xr + x25 в ?2+rCr в xr + x20 в ?2+rCr в xr – x31 в ?2+rCr в xr
= ?2+rCr в xr – ?2+rCr в xr+11 – ?2+rCr в xr+6 + ?2+rCr в xr+17 – ?2+rCr в xr+14 + ?2+rCr в xr+25 + ?2+rCr в xr+20 – ?2+rCr в xr+25
Now we have to find co-efficeient of x15
= 2+15C152+4C42+9C92+1C1 (rest all terms have greater than x15, so its coefficients are 0)
= 17C156C411C93C1
= 17C26C211C23C1
= {(17в16)/2} – {(6в5)/2} – {(11в10)/2} – 3
= (17в8) – (3в5) – (11в5) – 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 в 63 = 1008

Question 8.
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

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 = 12C35C3 = 220 – 10 = 210

Question 9.
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

Hint:
The number of combination of n distinct objects taken r at a time be x is given by
nCr = n!/{(n – r)! в 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 xв(r!).
Consequently, the number of permutations of n things, taken r at a time is xв(r!) and it is equal to nPr
So, xв(r!) = nPr
? xв(r!) = n!/(n – r)!
? x = n!/{(n – r)! в r!}
? nCr = n!/{(n – r)! в r!}

Question 10.
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

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 в 10 в 9 = 450

Question 11.
Let Tn denote the number of triangles which can be formed using the vertices of a regular polygon on n sides. If Tn+1 – Tn = 21, then n equals
(a) 5
(b) 7
(c) 6
(d) 4

Hint:
The number of triangles that can be formed using the vertices of a regular polygon = nC3
Given, Tn+1 – Tn = 21
? n+1C3nC3 = 21
? nC2 + nC3nC3 = 21 {since n+1Cr = nCr-1 + nCr}
? nC2 = 21
? n(n – 1)/2 = 21
? n(n – 1) = 21в2
? n╡ – n = 42
? n╡ – n – 42 = 0
? (n – 7)в(n + 6) = 0
? n = 7, -6
Since n can not be negative,
So, n = 7

Question 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

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

Question 13.
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

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 в9 в8
= 720

Question 14.
How many factors are 25 в 36 в 52 are perfect squares
(a) 24
(b) 12
(c) 16
(d) 22

Hint:
Any factors of 25 в 36 в 52 which is a perfect square will be of the form 2a в 3b в 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 в 4 в 2 = 24

Question 15.
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

Hint:
There are two cases
1. When 4 is selected from the first 5 and rest 6 from remaining 8
Total arrangement = 5C4 в 8C6
= 5C1 в 8C2
= 5 в (8в7)/(2в1)
= 5 в 4 в 7
= 140
2. When all 5 is selected from the first 5 and rest 5 from remaining 8
Total arrangement = 5C5 в 8C5
= 1 в 8C3
= (8в7в6)/(3в2в1)
= 8в7
= 56
Now, total number of choices available = 140 + 56 = 196

Question 16.
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

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.
= (6в6в6в6) – (5в5в5в5)
= 1296 – 625
= 671

Question 17.
In how many ways in which 8 students can be sated in a line is
(a) 40230
(b) 40320
(c) 5040
(d) 50400

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

Question 18.
The number of squares that can be formed on a chess board is
(a) 64
(b) 160
(c) 224
(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 = 8╡
Similarly, the number of squares with 2, 3, ….. 8 unit side will be 7╡, 6╡, …… 1╡
Hence, total number of square = 8╡ + 7╡ + ……+ 1╡ = 204

Question 19.
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

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.
= (6в6в6в6) – (5в5в5в5)
= 1296 – 625
= 671

Question 20.
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

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 в 10 в 9 = 450

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