MCQGeeks
0 : 0 : 1
CBSE
JEE
NTSE
NEET
English
UK Quiz
Quiz
Driving Test
Practice
Games
CBSE
Class 11 Maths
Permutations And Combinations
Quiz 1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
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
0%
604800
0%
17280
0%
120960
0%
518400
Explanation
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
7
P
4
=
7
P
3
= 7!/3! = (7 × 6 × 5 × 4 × 3!)/3!
= 7 × 6 × 5 × 4 = 840
Now, total number of arrangement = 6! × 840
= 720 × 840
= 604800
There are 12 points in a plane out of which 5 are collinear. The number of triangles formed by the points as vertices is
0%
185
0%
210
0%
220
0%
175
Explanation
210
Hint:
Total number of triangles that can be formed with 12 points (if none of them are collinear)
=
12
C
3
(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
5
C
3
triangles from the above count.
Hence, required number of triangles =
12
C
3
-
5
C
3
= 220 - 10 = 210
The number of combination of n distinct objects taken r at a time be x is given by
0%
n/2
C
r
0%
n/2
C
r/2
0%
n
C
r/2
0%
<
n
C
r
Explanation
n
C
r
Hint:
The number of combination of n distinct objects taken r at a time be x is given by
n
C
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
n
P
r
So, x×(r!) =
n
P
r
⇒ x×(r!) = n!/(n - r)!
⇒ x = n!/{(n - r)! × r!}
⇒
n
C
r
= n!/{(n - r)! × r!}
If repetition of the digits is allowed, then the number of even natural numbers having three digits is
0%
250
0%
350
0%
450
0%
550
Explanation
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
The number of ways in which 8 distinct toys can be distributed among 5 children is
0%
5
8
0%
8
5
0%
8
P
5
0%
5
P
5
Explanation
5
8
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
= 5
8
The value of P(n, n - 1) is
0%
n
0%
2n
0%
n!
0%
2n!
Explanation
n!
Hint:
Given,
Given, P(n, n - 1)
= n!/{(n - (n - 1)}
= n!/(n - n + 1)}
= n!
So, P(n, n - 1) = n!
In how many ways can 4 different balls be distributed among 5 different boxes when any box can have any number of balls?
0%
5
8
0%
8
5
0%
8
P
5
0%
5
P
5
Explanation
5
8
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
= 5
8
The number of ways of painting the faces of a cube with six different colors is
0%
1
0%
6
0%
6!
0%
None of these
Explanation
1
Hint:
Since the number of faces is same as the number of colors,
therefore the number of ways of painting them is 1
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
0%
1800
0%
1080
0%
1008
0%
8001
Explanation
1008
Hint:
Given there are 5 apples, 10 mangoes and 13 oranges.
Let x
1
is for apple, x
2
is for mango and x
3
is for orange.
Now, first we have to select total 15 fruits out of them.
x
1
+ x
2
+ x
3
= 15 (where 0 ⇐ x
1
⇐ 5, 0 ⇐ x
2
⇐ 10, 0 ⇐ x
3
⇐ 13)
= (x
0
+ x
1
+ x
2
+.........+ x
5
)×(x
0
+ x
1
+ x
2
+.........+ x1
10
)×(x
0
+ x
1
+ x
2
+.........+ x
13
)
= {(1- x
6
)/(1 - x)}×{(1- x
11
)/(1 - x)}×{(1- x
14
)/(1 - x)}
= {(1- x
6
)×(1- x
11
)×{(1- x
14
)}/(1 - x)³
= {(1- x
6
)×(1- x
11
)×{(1- x
14
)} × ∑
3+r+1
C
r
× x
r
= {(1- x
11
- x
6
+ x
17
)×{(1- x
14
)} × ∑
3+r+1
C
r
× x
r
= {(1- x
11
- x
6
+ x
17
- x
14
+ x
25
+ x
20
- x
31
)} × ∑
2+r
C
r
× x
r
= 1 × ∑
2+r
C
r
× x
r
- x
11
× ∑
2+r
C
r
× x
r
- x
6
× ∑
2+r
C
r
× x
r
+ x
17
× ∑
2+r
C
r
× x
r
- x
14
× ∑
2+r
C
r
× x
r
+ x
25
× ∑
2+r
C
r
× x
r
+ x
20
× ∑
2+r
C
r
× x
r
- x
31
× ∑
2+r
C
r
× x
r
= ∑
2+r
C
r
× x
r
- ∑
2+r
C
r
× x
r+11
- ∑
2+r
C
r
× x
r+6
+ ∑
2+r
C
r
× x
r+17
- ∑
2+r
C
r
× x
r+14
+ ∑
2+r
C
r
× x
r+25
+ ∑
2+r
C
r
× x
r+20
- ∑
2+r
C
r
× x
r+25
Now we have to find co-efficeient of x
15
=
2+15
C
15
-
2+4
C
4
-
2+9
C
9
-
2+1
C
1
(rest all terms have greater than x
15
, so its coefficients are 0)
=
17
C
15
-
6
C
4
-
11
C
9
-
3
C
1
=
17
C
2
-
6
C
2
-
11
C
2
-
3
C
1
= {(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 x
1
+ x
2
= 15
=
2-1+15
C
15
=
16
C
15
=
16
C
1
= 16
Now total number of ways of distribution = 16 × 63 = 1008
The number of ways can the letters of the word ASSASSINATION be arranged so that all the S are together is
0%
152100
0%
1512
0%
15120
0%
151200
Explanation
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
How many ways are here to arrange the letters in the word GARDEN with the vowels in alphabetical order?
0%
120
0%
240
0%
360
0%
480
Explanation
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
How many factors are 2
5
× 3
6
× 5
2
are perfect squares
0%
24
0%
12
0%
16
0%
22
Explanation
4
Hint:
Any factors of 2
5
× 3
6
× 5
2
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
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
0%
40
0%
196
0%
280
0%
346
Explanation
196
Hint:
There are two cases
1. When 4 is selected from the first 5 and rest 6 from remaining 8
Total arrangement =
5
C
4
×
8
C
6
=
5
C
1
×
8
C
2
= 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 =
5
C
5
×
8
C
5
= 1 ×
8
C
3
= (8×7×6)/(3×2×1)
= 8×7
= 56
Now, total number of choices available = 140 + 56 = 196
Four dice are rolled. The number of possible outcomes in which at least one dice show 2 is
0%
1296
0%
671
0%
625
0%
585
Explanation
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
In how many ways in which 8 students can be sated in a line is
0%
40230
0%
40320
0%
5040
0%
50400
Explanation
40320
Hint:
The number of ways in which 8 students can be sated in a line =
8
P
8
= 8!
= 40320
The number of squares that can be formed on a chess board is
0%
64
0%
160
0%
224
0%
204
Explanation
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 <sup>8</sup>C<sub>1</sub> = 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
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
0%
720
0%
420
0%
none of these
0%
5040
Explanation
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
=
10
P
3
= 10 ×9 ×8
= 720
0 h : 0 m : 1 s
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
0
Answered
0
Not Answered
0
Not Visited
Correct : 0
Incorrect : 0
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Support mcqgeeks.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page