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MCQ Class 11 Mathematics Permutations and Combinations with Answers - Set - 2
Question 1:
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
Correct Answer – (A)
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
Question 2 :
In how many ways in which 8 students can be sated in a line is
(a) 40230
(b) 40320
(c) 5040
(d) 50400
Correct Answer – (B)
Question 3 :
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
Correct Answer – (B)
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
Question 4 :
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
Correct Answer – (C)
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 5 :
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
MCQ Class 11 Mathematics Permutations and Combinations with Answers
Question 6 :
The number of squares that can be formed on a chess board is
(a) 64
(b) 160
(c) 224
(d) 204
Correct Answer – (D)
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
Question 7 :
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 8 :
How many factors are 25 × 36 × 52 are perfect squares
(a) 24
(b) 12
(c) 16
(d) 22
Correct Answer – (A)
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
Question 9 :
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
Correct Answer – (B)
The number of triangles that can be formed using the vertices of a regular polygon = nC3
Given, Tn+1 – Tn = 21
⇒ n+1C3 – nC3 = 21
⇒ nC2 + nC3 – nC3 = 21 {since n+1Cr = nCr-1 + nCr}
⇒ 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
Question 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
Correct Answer – (D)
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
- 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
