This set of Aptitude Questions and Answers (MCQs) focuses on “Permutations and Combinations – Set 4”.
1. When two dice are rolled at a time, what is the probability that one of them will be 4?
a) 11 / 36
b) 10 / 36
c) 4!
d) 8 / 36
View Answer
Explanation: When two dice are rolled at a time the possible outcomes will be = 36.
The outcomes when one of them will be 4 are: (1,4), (2,4), (3,4), (4,4), (5,4), (6,4), (4,1), (4,2), (4,3), (4,5), (4,6) = 11.
The probability of getting one 4 is = 11 / 36.
2. How many squares can be formed from square of 4 rows and 4 columns?
a) 12
b) 16
c) 28
d) 30
View Answer
Explanation: For a square of 1 row and 1 column number of squares will be = 12 = 1.
For a square of 2 rows and 2 columns number of squares will be = 12 + 22 = 1 + 4 = 5.
For a square of 3 rows and 3 columns number of squares will be = 12 + 22 + 32 = 1 + 4 + 9 = 14.
For a square of n rows and n columns, number of squares will be = 12 + 22 + 32 + ……….. + n2.
Here, we have 4 rows and 4 columns.
Therefore, n = 4 ⇒ 12 + 22 + 32 + 42 = 1 + 4 + 9 + 16 = 30.
Number of squares can be formed from a square of 4rows and 4 columns = 30.
3. In how many ways a group of 4 boys and 4 girls can be made out of 8 boys and 6 girls?
a) 1250
b) 720
c) 1050
d) 360
View Answer
Explanation: Given, 4 out of 8 boys and 4 out of 6 girls are to be selected.
So, (nCr * nCr). Where nCr = n! / r!(n – r)!.
(8C4 * 6C4) = (8! / 4!(8 – 4)! * 6! / 4!(6 – 4)!).
(8 * 7 * 6 * 5 * 4!) / (4! * 4!) * (6 * 5 * 4!) / (4! * 2!))
(8 * 7 * 6 * 5 * 4!) / (4! * 4!) = 70
(6 * 5 * 4!) / (4! * 2!)) = 15
Number of ways required = 70 * 15 = 1050.
4. There are 14 points on a plane of which 4 are collinear. Find the number of straight lines that can be drawn using these points.
a) 10!
b) 14!
c) 84
d) 98
View Answer
Explanation: 14 points in a plane of which 4 are collinear.
We can find this using nCr = n! / r!(n – r)!.
A straight line a can be formed using any points in the plane.
So, number of points in the plane is denoted by n = 14.
Two points are taken to form a straight line r = 2.
The 4 collinear points are already included in the 14 points.
In order to eliminate the double counting we use 14C2 – 4C2.
There is one common line that joins all the 4 collinear points.
So, number of straight lines that can be drawn = 14C2 – 4C2 + 1 = 91 + 6 + 1 = 98.
5. From 4 vowels and 8 consonants, how many words can be formed using 2 vowels and 3 consonants?
a) 16480
b) 25200
c) 10080
d) 46800
View Answer
Explanation: Given that, 2 out of 4 vowels and 3 out of 8 consonants.
So, (nCr * nCr). Where nCr = n! / r!(n – r)!.
(4C2 * 8C3) = (4! / 2!(4 – 2)! * 8! / 2!(8 – 3)!) = (4 * 3 * 2!) / (2! * 2!) * (8 * 7 * 6 * 5!) / (4! * 5!)).
(6 * 14) = 84.
Number of words, each having 2 vowels and 3 consonants = 84.
Each word contains 5 letters.
Number of ways of arranging 5 letters among themselves = 5! = 120.
Required number of ways = (84 * 120) = 10080.
6. How many different words can be formed from the word ‘REDUCTION’ which start with ‘D’ and end with ‘R’?
a) 9! / 7!
b) 63
c) 7!
d) 9!
View Answer
Explanation: The word ‘REDUCTION’ has 9 letters.
The words should start with D and end with R.
We can write it as D_ _ _ _ _ _ _R.
Remaining seven letters are E, U, C, T, I, O, N.
The letters are distinct, so that we can arrange them in 7! ways.
Number of different words that can be formed = 7!.
7. In a college 5 Engineering students and 5 Medicine students are discussing about the requirements in their labs. In how many ways can we arrange them so that no two Engineering students should sit together?
a) 5! * 6P5
b) 5! * 5!
c) 10080
d) 5! * 4!
View Answer
Explanation: Given that, there are 5 Engineering students and 5 Medicine students.
First of all, all the Engineering students should arrange themselves alternately in 5! ways = 120.
So, 5 Medicine students are left with 6 places.
Medicine students can arrange themselves in 6P5 ways.
Therefore, total number of ways will be = 5! * 6P5.
8. In how many ways 10 red balls can be chosen out 50 red balls?
a) 10!
b) 1
c) 50! / 10!
d) 500
View Answer
Explanation: Given that, we have to choose 10 balls out 50 balls.
As all the 50 balls are of red color, it means they are identical.
We have to select 10 red balls.
Number of ways to select 10 red balls out of 50 = 1.
9. nCr = r! * nPr.
a) True
b) False
View Answer
Explanation: Formula for nCr = n! / r!(n – r)!.
nPr = n! / (n – r)!
So, n! / r!(n – r)! = n! / (n – r)!.
nPr = r! * nCr
nCr = nPr / r!
10. Find the sum of number of sides and number of diagonals of a Decagon?
a) 45
b) 10!
c) 10! / 2!
d) 20
View Answer
Explanation: Decagon has 10 vertices.
A line can be formed by joining any two vertices.
So, the sum of number of sides and number of diagonals of a Decagon is given by nCr.
The formula for nCr = n! / r!(n – r)!.
Number of vertices of a Decagon n = 10.
Number of points taken at a time r = 2.
10C2 = 45
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