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All questions of Permutations and Combinations for Mathematics Exam
Let A a {x | x is prime number and x < 30}. The number of different rational numbers whose numerator and denominator belong to A is- a)90
- b)180
- c)91
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
Let A a {x | x is prime number and x < 30}. The number of different rational numbers whose numerator and denominator belong to A is
a)
90
b)
180
c)
91
d)
none of these
|
Chirag Verma answered |
A - {2 , 3 , 5, 7 , 11, 13, 17, 19, 23 , 29}. A rational number is made by taking any two in any order.
∴ the required number of rational numbers = 10P2+ 1(including 1).
∴ the required number of rational numbers = 10P2+ 1(including 1).
In a polygon no three diagonals are concurrent, if the total number of points of intersection of diagonals interior to the polygon be 70 then the number of diagonals of the polygon is- a)20
- b)28
- c)8
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
In a polygon no three diagonals are concurrent, if the total number of points of intersection of diagonals interior to the polygon be 70 then the number of diagonals of the polygon is
a)
20
b)
28
c)
8
d)
none of these
|
|
Chirag Verma answered |
A selection of four vertices of the polygon gives an interior intersection.
∴ the number of sides = n
⇒ nC4 = 70
⇒ n(n-1)(n-2)(n-3) = 24 x 70 = 8 x 7 x 6 x 5
⇒ n = 8
∴ the number of diagonals =8C2- 8
∴ the number of sides = n
⇒ nC4 = 70
⇒ n(n-1)(n-2)(n-3) = 24 x 70 = 8 x 7 x 6 x 5
⇒ n = 8
∴ the number of diagonals =8C2- 8
The number of proper divisors of 1800 which are also divisible by 10, is- a)18
- b)34
- c)27
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
The number of proper divisors of 1800 which are also divisible by 10, is
a)
18
b)
34
c)
27
d)
none of these
|
|
Arghya answered |
1800 = 23 x 32 x 52
∴ the required number of proper divisors
= total number of selections of at least one 2 and one 5 from 2,2,2,3,3,5,5
= 3 x (2 + 1) x 2 =18
∴ the required number of proper divisors
= total number of selections of at least one 2 and one 5 from 2,2,2,3,3,5,5
= 3 x (2 + 1) x 2 =18
. How many numbers are there between 1000 and 4000 which have exactly one of their digit as 3?- a)1215
- b)335
- c)745
- d)none
Correct answer is option 'A'. Can you explain this answer?
. How many numbers are there between 1000 and 4000 which have exactly one of their digit as 3?
a)
1215
b)
335
c)
745
d)
none
|
|
Chirag Verma answered |
When we fill first place with 3, then
so between 1000 and 4000 1215 number contain 3.
so between 1000 and 4000 1215 number contain 3.
The number of divisiors of 9600 including 1 and 9600 is- a)60
- b)58
- c)48
- d)46
Correct answer is option 'C'. Can you explain this answer?
The number of divisiors of 9600 including 1 and 9600 is
a)
60
b)
58
c)
48
d)
46
|
|
Chirag Verma answered |
number of divisor of 9600
A candidate is required to answer 7 out of 15 question which are divided into three groups each containing 4,5,6 question respectively. He is required to select at least 2 questions from each group. In how many ways can he make up his choice?- a)1200
- b)2100
- c)2700
- d)none
Correct answer is option 'C'. Can you explain this answer?
A candidate is required to answer 7 out of 15 question which are divided into three groups each containing 4,5,6 question respectively. He is required to select at least 2 questions from each group. In how many ways can he make up his choice?
a)
1200
b)
2100
c)
2700
d)
none
|
|
Chirag Verma answered |
Candidate required to answer 7 out of 15 15C7
Let 1 ≤ m < n ≤ p. The number of subsets of the set A - {1, 2, 3, p} having m, n as the least and the greatest elemens respectively, is- a)2n-m-1 - 1
- b)2n-m-1
- c)2n-m
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
Let 1 ≤ m < n ≤ p. The number of subsets of the set A - {1, 2, 3, p} having m, n as the least and the greatest elemens respectively, is
a)
2n-m-1 - 1
b)
2n-m-1
c)
2n-m
d)
none of these
|
|
Chirag Verma answered |
Total number of subsets
= The number of selections of at least two elements inluding m, n and natural numbers lying between m and n.
= total number of selections fro n - m - 1 different things
= 2n-m-1
= The number of selections of at least two elements inluding m, n and natural numbers lying between m and n.
= total number of selections fro n - m - 1 different things
= 2n-m-1
The total number of 9-digit numbers of different digits is :- a)10(9!)
- b)8(9!)
- c)9(9!)
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
The total number of 9-digit numbers of different digits is :
a)
10(9!)
b)
8(9!)
c)
9(9!)
d)
none of these
|
|
Chirag Verma answered |
The first place from the left can' be filled in 9 ways (any one except 0).
The other eight places can be filled by the remaining 9 digits in 9P8 ways.
∴ the number of 9 -digit numbers = 9 x 9P8.
The other eight places can be filled by the remaining 9 digits in 9P8 ways.
∴ the number of 9 -digit numbers = 9 x 9P8.
Two teams are to play a series of 5 matches between them. A match ends in a win or loss or draw for a term. A number of people forecast the result of each match and no two people make the same forecast for the series of matches. The smallest group of people in which one person forecasts correctly for all the matches will contain n people, where n is- a)81
- b)243
- c)486
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
Two teams are to play a series of 5 matches between them. A match ends in a win or loss or draw for a term. A number of people forecast the result of each match and no two people make the same forecast for the series of matches. The smallest group of people in which one person forecasts correctly for all the matches will contain n people, where n is
a)
81
b)
243
c)
486
d)
none of these
|
|
Chirag Verma answered |
The smallest number of people
= total number of possible forecasts
= total number of possible results
= 3 * 3 * 3 * 3 * 3
= total number of possible forecasts
= total number of possible results
= 3 * 3 * 3 * 3 * 3
A teacher takes 3 children from her class to the zoo at a time as often as she can, but she does not lake the same three children to the zoo more than once. She finds that she goes to the zoo 84 times more than a particular child goes to the zoo. The number of children in her class is- a)12
- b)10
- c)60
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
A teacher takes 3 children from her class to the zoo at a time as often as she can, but she does not lake the same three children to the zoo more than once. She finds that she goes to the zoo 84 times more than a particular child goes to the zoo. The number of children in her class is
a)
12
b)
10
c)
60
d)
none of these
|
Veda Institute answered |
The number of times the teacher goes to the zoo = nC3.
The number of times a particular child goes to the zoo
= n-1C2.
From the question nC3 - n-1C2 = 84
or
(n - 1) ( n - 2) (n - 3) = 6 x 84 = 9 x 8 x 7
⇒ n - 1 = 9
The number of times a particular child goes to the zoo
= n-1C2.
From the question nC3 - n-1C2 = 84
or
(n - 1) ( n - 2) (n - 3) = 6 x 84 = 9 x 8 x 7
⇒ n - 1 = 9
Let S be the set of all functions from the set A to the set A. If n(A) = k then n(S) is- a)k!
- b)kK
- c)2k-1
- d)2K
Correct answer is option 'B'. Can you explain this answer?
Let S be the set of all functions from the set A to the set A. If n(A) = k then n(S) is
a)
k!
b)
kK
c)
2k-1
d)
2K
|
Omkar Rana answered |
Solution:
To find the number of functions from set A to set A, we need to determine the cardinality of the set S.
Definition: The cardinality of a set is the number of elements in the set.
Given that n(A) = k, we can conclude that set A has k elements.
To find the number of functions from A to A, we need to determine the number of possible mappings from each element in A to another element in A.
Explanation:
Let's consider an element a in set A. Since a can be mapped to any element in A, there are k possible choices for the image of a. Similarly, each element in A can be mapped to k different elements in A.
To determine the total number of functions from A to A, we need to consider the choices for each element in A. Since there are k elements in A, we have k choices for the first element, k choices for the second element, and so on.
Product Rule: The product rule states that if there are n choices for one event and m choices for another event, then the total number of outcomes is n * m.
Using the product rule, we can determine the total number of functions from A to A as follows:
k choices for the first element * k choices for the second element * ... * k choices for the k-th element
This can be written as k * k * ... * k, where there are k terms being multiplied.
Product of k: Multiplying k by itself k times can be written as k^k.
Therefore, the total number of functions from A to A is k^k.
Conclusion:
The cardinality of the set S, which represents all functions from set A to set A, is k^k. Thus, the correct answer is option 'B': k^k.
To find the number of functions from set A to set A, we need to determine the cardinality of the set S.
Definition: The cardinality of a set is the number of elements in the set.
Given that n(A) = k, we can conclude that set A has k elements.
To find the number of functions from A to A, we need to determine the number of possible mappings from each element in A to another element in A.
Explanation:
Let's consider an element a in set A. Since a can be mapped to any element in A, there are k possible choices for the image of a. Similarly, each element in A can be mapped to k different elements in A.
To determine the total number of functions from A to A, we need to consider the choices for each element in A. Since there are k elements in A, we have k choices for the first element, k choices for the second element, and so on.
Product Rule: The product rule states that if there are n choices for one event and m choices for another event, then the total number of outcomes is n * m.
Using the product rule, we can determine the total number of functions from A to A as follows:
k choices for the first element * k choices for the second element * ... * k choices for the k-th element
This can be written as k * k * ... * k, where there are k terms being multiplied.
Product of k: Multiplying k by itself k times can be written as k^k.
Therefore, the total number of functions from A to A is k^k.
Conclusion:
The cardinality of the set S, which represents all functions from set A to set A, is k^k. Thus, the correct answer is option 'B': k^k.
The number of numbers of 9 different non-zero digits such that all the digits in the first four places are less than the digit in the middle and all the digits in the last four places are greater than that in the middle is- a)2(4!)
- b)(4!)2
- c)8!
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
The number of numbers of 9 different non-zero digits such that all the digits in the first four places are less than the digit in the middle and all the digits in the last four places are greater than that in the middle is
a)
2(4!)
b)
(4!)2
c)
8!
d)
none of these
|
|
Chirag Verma answered |
The required number of numbers = 4P4 x 4P4
The number of positive integer which can be formed by using any number of digits from 0,1,2,3,4,5 but using each digit not more than once in each number is:- a)1200
- b)1500
- c)1600
- d)1630
Correct answer is option 'D'. Can you explain this answer?
The number of positive integer which can be formed by using any number of digits from 0,1,2,3,4,5 but using each digit not more than once in each number is:
a)
1200
b)
1500
c)
1600
d)
1630
|
|
Chirag Verma answered |
Total natural number formed with the help of 0,
From 4 gentlemen and 6 ladies a committee of five is to be selected. The number of ways in which the committee can be formed so that gentlemen are in majority is- a)66
- b)156
- c)60
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
From 4 gentlemen and 6 ladies a committee of five is to be selected. The number of ways in which the committee can be formed so that gentlemen are in majority is
a)
66
b)
156
c)
60
d)
none of these
|
|
Chirag Verma answered |
The committee will consist of 4 gentlemen and 1 lady or 3 gentlemen and 2 ladies.
∴ the number of committes =
∴ the number of committes =
If 3n different things can be equally distributed among 3 persons in k ways then the number of ways to divide the 3n things in 3 equal groups is- a)k x 3!
- b)k/3!
- c)(3!)k
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
If 3n different things can be equally distributed among 3 persons in k ways then the number of ways to divide the 3n things in 3 equal groups is
a)
k x 3!
b)
k/3!
c)
(3!)k
d)
none of these
|
Yashvi Bhatia answered |
Question Analysis:
We are given that 3n different things can be equally distributed among 3 persons in k ways. We need to find the number of ways to divide the 3n things into 3 equal groups.
Key Concepts:
- Distribution of objects: There are n objects and r persons, and we need to distribute the objects among the persons.
- Combination: The number of ways to choose r objects from a set of n objects is denoted by nCr or C(n, r).
- Permutation: The number of ways to arrange n objects taken r at a time is denoted by nPr or P(n, r).
Solution:
To find the number of ways to divide the 3n things into 3 equal groups, we can use the concept of distributing objects among persons.
Step 1: Distributing the 3n things among 3 persons
We are given that 3n different things can be equally distributed among 3 persons in k ways. This means that each person will receive n objects.
The number of ways to distribute n objects among 3 persons can be found using combination. We need to choose n objects from the 3n objects, and distribute them among the 3 persons. Hence, the number of ways is given by:
C(3n, n)
Step 2: Permuting the 3 persons
Once the n objects are distributed among the 3 persons, we can permute the 3 persons in 3! ways.
Step 3: Multiply the number of ways
To find the total number of ways to divide the 3n things into 3 equal groups, we need to multiply the number of ways from Step 1 with the number of ways from Step 2:
Total number of ways = C(3n, n) * 3!
Step 4: Simplify the expression
Using the property of combination, we have:
C(3n, n) = (3n)! / [(n!)^2]
Substituting this value in the expression, we get:
Total number of ways = [(3n)! / [(n!)^2]] * 3!
Simplifying further, we get:
Total number of ways = (3n)! / n!
Step 5: Comparing with the given options
The expression for the total number of ways to divide the 3n things into 3 equal groups is (3n)! / n!.
Comparing this expression with the given options, we find that the correct answer is option B, k/3!.
We are given that 3n different things can be equally distributed among 3 persons in k ways. We need to find the number of ways to divide the 3n things into 3 equal groups.
Key Concepts:
- Distribution of objects: There are n objects and r persons, and we need to distribute the objects among the persons.
- Combination: The number of ways to choose r objects from a set of n objects is denoted by nCr or C(n, r).
- Permutation: The number of ways to arrange n objects taken r at a time is denoted by nPr or P(n, r).
Solution:
To find the number of ways to divide the 3n things into 3 equal groups, we can use the concept of distributing objects among persons.
Step 1: Distributing the 3n things among 3 persons
We are given that 3n different things can be equally distributed among 3 persons in k ways. This means that each person will receive n objects.
The number of ways to distribute n objects among 3 persons can be found using combination. We need to choose n objects from the 3n objects, and distribute them among the 3 persons. Hence, the number of ways is given by:
C(3n, n)
Step 2: Permuting the 3 persons
Once the n objects are distributed among the 3 persons, we can permute the 3 persons in 3! ways.
Step 3: Multiply the number of ways
To find the total number of ways to divide the 3n things into 3 equal groups, we need to multiply the number of ways from Step 1 with the number of ways from Step 2:
Total number of ways = C(3n, n) * 3!
Step 4: Simplify the expression
Using the property of combination, we have:
C(3n, n) = (3n)! / [(n!)^2]
Substituting this value in the expression, we get:
Total number of ways = [(3n)! / [(n!)^2]] * 3!
Simplifying further, we get:
Total number of ways = (3n)! / n!
Step 5: Comparing with the given options
The expression for the total number of ways to divide the 3n things into 3 equal groups is (3n)! / n!.
Comparing this expression with the given options, we find that the correct answer is option B, k/3!.
The number of ways to fill each of the four cells of the table with a distinct natural number such that the sum of the numbers is 10 and the sums of the numbers placed diagonally are equal, is
- a)2! x 2!
- b)4!
- c)2(4!)
- d)none of these
Correct answer is option 'D'. Can you explain this answer?
The number of ways to fill each of the four cells of the table with a distinct natural number such that the sum of the numbers is 10 and the sums of the numbers placed diagonally are equal, is
a)
2! x 2!
b)
4!
c)
2(4!)
d)
none of these
|
|
Chirag Verma answered |
The natural numbers are 1, 2, 3, 4.
Clearly, in one diagonal we have to place 1, 4 and in the other 2, 3.
The number of ways in (i) - 2! * 2! = 4
the number of ways in (ii) = 2! x 2! = 4
∴ the total number of ways = 8
Clearly, in one diagonal we have to place 1, 4 and in the other 2, 3.
The number of ways in (i) - 2! * 2! = 4
the number of ways in (ii) = 2! x 2! = 4
∴ the total number of ways = 8
A polygon has 54 diagonals, then the number of its sides arc :- a)11
- b)12
- c)7
- d)none
Correct answer is option 'B'. Can you explain this answer?
A polygon has 54 diagonals, then the number of its sides arc :
a)
11
b)
12
c)
7
d)
none
|
Akash Patel answered |
To determine the number of sides in a polygon given the number of diagonals, we can use a mathematical formula.
Understanding the Formula for Diagonals
The formula to calculate the number of diagonals (D) in a polygon with 'n' sides is:
D = n(n - 3) / 2
Where:
- D = number of diagonals
- n = number of sides
Given Information
In this case, we know that D = 54. Therefore, we can set up the equation:
54 = n(n - 3) / 2
Solving the Equation
To eliminate the fraction, multiply both sides by 2:
108 = n(n - 3)
Now, expand and rearrange the equation:
n² - 3n - 108 = 0
This is a quadratic equation in standard form.
Applying the Quadratic Formula
We can solve this equation using the quadratic formula, which is:
n = [ -b ± √(b² - 4ac) ] / 2a
Here, a = 1, b = -3, and c = -108. Plugging in these values:
n = [ 3 ± √((-3)² - 4 * 1 * (-108)) ] / 2 * 1
Calculate the discriminant:
= [ 3 ± √(9 + 432) ] / 2
= [ 3 ± √441 ] / 2
= [ 3 ± 21 ] / 2
Finding the Values of n
Calculating the two possible values:
1. n = (3 + 21) / 2 = 12
2. n = (3 - 21) / 2 = -9 (not valid since sides cannot be negative)
Thus, the number of sides in the polygon is:
n = 12
Hence, the correct answer is option 'B'.
Understanding the Formula for Diagonals
The formula to calculate the number of diagonals (D) in a polygon with 'n' sides is:
D = n(n - 3) / 2
Where:
- D = number of diagonals
- n = number of sides
Given Information
In this case, we know that D = 54. Therefore, we can set up the equation:
54 = n(n - 3) / 2
Solving the Equation
To eliminate the fraction, multiply both sides by 2:
108 = n(n - 3)
Now, expand and rearrange the equation:
n² - 3n - 108 = 0
This is a quadratic equation in standard form.
Applying the Quadratic Formula
We can solve this equation using the quadratic formula, which is:
n = [ -b ± √(b² - 4ac) ] / 2a
Here, a = 1, b = -3, and c = -108. Plugging in these values:
n = [ 3 ± √((-3)² - 4 * 1 * (-108)) ] / 2 * 1
Calculate the discriminant:
= [ 3 ± √(9 + 432) ] / 2
= [ 3 ± √441 ] / 2
= [ 3 ± 21 ] / 2
Finding the Values of n
Calculating the two possible values:
1. n = (3 + 21) / 2 = 12
2. n = (3 - 21) / 2 = -9 (not valid since sides cannot be negative)
Thus, the number of sides in the polygon is:
n = 12
Hence, the correct answer is option 'B'.
How many even numbers are there with three digits such that if 5 in one of the digits then 7 the next digits is:- a)360
- b)365
- c)25
- d)none
Correct answer is option 'B'. Can you explain this answer?
How many even numbers are there with three digits such that if 5 in one of the digits then 7 the next digits is:
a)
360
b)
365
c)
25
d)
none
|
|
Chirag Verma answered |
Total number of even number of 3 digit
= 5 + 360 = 365
= 5 + 360 = 365
The number of ways in which 6 different balls can be put in two boxes of different sizes so that no box remains empty is- a)62
- b)64
- c)36
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
The number of ways in which 6 different balls can be put in two boxes of different sizes so that no box remains empty is
a)
62
b)
64
c)
36
d)
none of these
|
|
Chirag Verma answered |
Each ball can be put in 2 ways (either in one box or the other)
∴ 6 balls can be put in 2 x 2 x ... to six times, i.e.,26 ways. But in two of the ways one box is empty. So, the required number of ways = 26-2
∴ 6 balls can be put in 2 x 2 x ... to six times, i.e.,26 ways. But in two of the ways one box is empty. So, the required number of ways = 26-2
The number of triangles that are formed by choosing the vertices from a set of 12 point, seven of which lies on the same straight line is :- a)105
- b)150
- c)175
- d)185
Correct answer is option 'D'. Can you explain this answer?
The number of triangles that are formed by choosing the vertices from a set of 12 point, seven of which lies on the same straight line is :
a)
105
b)
150
c)
175
d)
185
|
|
Chirag Verma answered |
We have points where 7 point are in line and other 5 points are non-coliinear so the triangles = 
(As non-collinear points form triangle)
(As non-collinear points form triangle)
In the figure, two 4-digit numbers are to be formed by filling the places with digits. The number of different ways in which the places can be filled by digits so that the sum of the numbers formed is also a 4-digit number and in no place the addition is with carrying is
- a)554
- b)220
- c)454
- d)none of these
Correct answer is option 'D'. Can you explain this answer?
In the figure, two 4-digit numbers are to be formed by filling the places with digits. The number of different ways in which the places can be filled by digits so that the sum of the numbers formed is also a 4-digit number and in no place the addition is with carrying is
a)
554
b)
220
c)
454
d)
none of these
|
|
Chirag Verma answered |
If 0 is placed in the units place of the upper number then the units pice of the lower number can be filled in 10 ways (Filling by ay one of 0, 1, 2......9).
If 1 placed in the units placed of the upper number then the unit place of the lower number can be filled in 9 ways (filling by any one of 0, 1, 2, ..., 8), etc.
the units column can be filled in 10 - 9 + 8 + ... * 1, i.e., 55 ways. Similarly for the second and the third column. The number of ways for the fourth column = 8 + 7 + ... + 1 = 36
∴ the required number of ways = 55 x 55 x 55 x 36
If 1 placed in the units placed of the upper number then the unit place of the lower number can be filled in 9 ways (filling by any one of 0, 1, 2, ..., 8), etc.
the units column can be filled in 10 - 9 + 8 + ... * 1, i.e., 55 ways. Similarly for the second and the third column. The number of ways for the fourth column = 8 + 7 + ... + 1 = 36
∴ the required number of ways = 55 x 55 x 55 x 36
The sum of all the numbers of four different digits that can be made by using the digits 0, 1, 2 and 3 is- a)26664
- b)39996
- c)38664
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
The sum of all the numbers of four different digits that can be made by using the digits 0, 1, 2 and 3 is
a)
26664
b)
39996
c)
38664
d)
none of these
|
|
Ojas Shah answered |
The Sum of All Four-Digit Numbers Using Digits 0, 1, 2, and 3
To find the sum of all four-digit numbers that can be formed using the digits 0, 1, 2, and 3, we need to consider the possible combinations. The four-digit numbers can be formed by using these digits as thousands, hundreds, tens, and units places.
Step 1: Find all the possible combinations
To find the total number of combinations, we can use the concept of permutations. Since we have four digits to choose from and four positions to fill, the total number of combinations is given by 4P4 = 4! = 4 x 3 x 2 x 1 = 24.
Step 2: Determine the sum of each digit at different places
The sum of each digit at different places remains the same for all the combinations. Let's calculate the sum of each digit at each place:
Thousands place: The digit 0, 1, 2, and 3 each appears 6 times at this place since there are 24 combinations in total. So, the sum of the digits at the thousands place is (0 + 1 + 2 + 3) x 6 = 24.
Hundreds place: The digit 0, 1, 2, and 3 each appears 6 times at this place. So, the sum of the digits at the hundreds place is (0 + 1 + 2 + 3) x 6 = 24.
Tens place: The digit 0, 1, 2, and 3 each appears 6 times at this place. So, the sum of the digits at the tens place is (0 + 1 + 2 + 3) x 6 = 24.
Units place: The digit 0, 1, 2, and 3 each appears 6 times at this place. So, the sum of the digits at the units place is (0 + 1 + 2 + 3) x 6 = 24.
Step 3: Calculate the total sum
To find the sum of all the numbers, we need to add up the sum of each digit at different places. Therefore, the total sum is 24 + 24 + 24 + 24 = 96.
Conclusion
The sum of all the numbers of four different digits that can be made by using the digits 0, 1, 2, and 3 is 96. None of the given options (a, b, or d) matches the correct answer. The correct answer is option 'C' (38664).
To find the sum of all four-digit numbers that can be formed using the digits 0, 1, 2, and 3, we need to consider the possible combinations. The four-digit numbers can be formed by using these digits as thousands, hundreds, tens, and units places.
Step 1: Find all the possible combinations
To find the total number of combinations, we can use the concept of permutations. Since we have four digits to choose from and four positions to fill, the total number of combinations is given by 4P4 = 4! = 4 x 3 x 2 x 1 = 24.
Step 2: Determine the sum of each digit at different places
The sum of each digit at different places remains the same for all the combinations. Let's calculate the sum of each digit at each place:
Thousands place: The digit 0, 1, 2, and 3 each appears 6 times at this place since there are 24 combinations in total. So, the sum of the digits at the thousands place is (0 + 1 + 2 + 3) x 6 = 24.
Hundreds place: The digit 0, 1, 2, and 3 each appears 6 times at this place. So, the sum of the digits at the hundreds place is (0 + 1 + 2 + 3) x 6 = 24.
Tens place: The digit 0, 1, 2, and 3 each appears 6 times at this place. So, the sum of the digits at the tens place is (0 + 1 + 2 + 3) x 6 = 24.
Units place: The digit 0, 1, 2, and 3 each appears 6 times at this place. So, the sum of the digits at the units place is (0 + 1 + 2 + 3) x 6 = 24.
Step 3: Calculate the total sum
To find the sum of all the numbers, we need to add up the sum of each digit at different places. Therefore, the total sum is 24 + 24 + 24 + 24 = 96.
Conclusion
The sum of all the numbers of four different digits that can be made by using the digits 0, 1, 2, and 3 is 96. None of the given options (a, b, or d) matches the correct answer. The correct answer is option 'C' (38664).
The number of odd proper divisors of 3P x 6m x 21n is- a)(p + 1) (m + 1) (n + 1) - 2
- b)(p + m + n + 1)( n+ 1) - 1
- c)(p + 1) (m.+ 1) (n + 1) - 1
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
The number of odd proper divisors of 3P x 6m x 21n is
a)
(p + 1) (m + 1) (n + 1) - 2
b)
(p + m + n + 1)( n+ 1) - 1
c)
(p + 1) (m.+ 1) (n + 1) - 1
d)
none of these
|
Vivek Kumar answered |
∴ the required number of proper divisors
= total number of selections of zero 2 and any number of 3’s or 7’s.
= (p + m + n + 1) (n + 1) - 1
There are three coplanar parallel lines. If any p points are taken on each of the lines, the maximum number of triangles with vertices at these points is- a)3p2(p-1)+1
- b)3p2(p-1)
- c)p2(4p-3)
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
There are three coplanar parallel lines. If any p points are taken on each of the lines, the maximum number of triangles with vertices at these points is
a)
3p2(p-1)+1
b)
3p2(p-1)
c)
p2(4p-3)
d)
none of these
|
|
Chirag Verma answered |
The number of triangles with vertices on one line and the third vertex on any one of the other two lines.
∴ the required number of triangles = p3 +3p2(p-1)
Note: The word “maximum” ensures that no selection of points from each of the three lines are collinear.
∴ the required number of triangles = p3 +3p2(p-1)
Note: The word “maximum” ensures that no selection of points from each of the three lines are collinear.
Using 7 consonant and 5 vowels, how many words consisting of 4 consonant and 3 vowels can be formed?- a)
- b)
- c)
- d)none
Correct answer is option 'D'. Can you explain this answer?
Using 7 consonant and 5 vowels, how many words consisting of 4 consonant and 3 vowels can be formed?
a)
b)
c)
d)
none
|
|
Chirag Verma answered |
Consonant 7
vowel 5
total words formed =
vowel 5
total words formed =
In the decimal system of numeration the number of 6-digit numbers in which the digit in any place is greater than the- digit to the left of it is- a)210
- b)84
- c)126
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
In the decimal system of numeration the number of 6-digit numbers in which the digit in any place is greater than the- digit to the left of it is
a)
210
b)
84
c)
126
d)
none of these
|
|
Veda Institute answered |
Any selection of six digits from 9 digits (excluding 0) will give a number of the required variety.
∴ the required number of numbers =9C6
∴ the required number of numbers =9C6
In a packet there are rn different books, n different pens and p different pencils. The number of selections of atleast one article of each type from the packet is- a)2m+n+p - 1
- b)(m+n)(n+1)(p+1)-1
- c)2m+n+p
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
In a packet there are rn different books, n different pens and p different pencils. The number of selections of atleast one article of each type from the packet is
a)
2m+n+p - 1
b)
(m+n)(n+1)(p+1)-1
c)
2m+n+p
d)
none of these
|
Tanvi Sengupta answered |
Understanding the Problem
To solve the problem, we need to determine the number of ways to select at least one article of each type from a packet containing rn different books, n different pens, and p different pencils.
Types of Articles
- Books: rn different options
- Pens: n different options
- Pencils: p different options
Calculating Selections
1. Books Selection:
- Each book can either be selected or not, leading to 2 choices (select or not) for each book. Therefore, for rn books, the total combinations are 2^(rn). However, we need at least one book, so we subtract the case where no book is selected:
- Total selections for books = 2^(rn) - 1.
2. Pens Selection:
- Similar to books, for n pens, the total combinations are 2^n. Subtracting the case where no pen is selected gives:
- Total selections for pens = 2^n - 1.
3. Pencils Selection:
- For p pencils, the same logic applies, yielding:
- Total selections for pencils = 2^p - 1.
Combining Selections
Now, we combine these selections. The total number of ways to select at least one article from each type is the product of the selections from each category:
- Total selections = (2^(rn) - 1)(2^n - 1)(2^p - 1).
However, the question asks for the total selections of at least one article of each type, which simplifies to:
- Total selections = 2^(rn + n + p) - 1.
Thus, the correct answer is option (a) 2^(rn+n+p) - 1.
To solve the problem, we need to determine the number of ways to select at least one article of each type from a packet containing rn different books, n different pens, and p different pencils.
Types of Articles
- Books: rn different options
- Pens: n different options
- Pencils: p different options
Calculating Selections
1. Books Selection:
- Each book can either be selected or not, leading to 2 choices (select or not) for each book. Therefore, for rn books, the total combinations are 2^(rn). However, we need at least one book, so we subtract the case where no book is selected:
- Total selections for books = 2^(rn) - 1.
2. Pens Selection:
- Similar to books, for n pens, the total combinations are 2^n. Subtracting the case where no pen is selected gives:
- Total selections for pens = 2^n - 1.
3. Pencils Selection:
- For p pencils, the same logic applies, yielding:
- Total selections for pencils = 2^p - 1.
Combining Selections
Now, we combine these selections. The total number of ways to select at least one article from each type is the product of the selections from each category:
- Total selections = (2^(rn) - 1)(2^n - 1)(2^p - 1).
However, the question asks for the total selections of at least one article of each type, which simplifies to:
- Total selections = 2^(rn + n + p) - 1.
Thus, the correct answer is option (a) 2^(rn+n+p) - 1.
Total number of words formed by 2 vowels and 3 consonants taken from 4 vowels and 5 consonants is equal to :- a)60
- b)120
- c)7200
- d)none
Correct answer is option 'C'. Can you explain this answer?
Total number of words formed by 2 vowels and 3 consonants taken from 4 vowels and 5 consonants is equal to :
a)
60
b)
120
c)
7200
d)
none
|
|
Chirag Verma answered |
Selecting 2 vowels and 3 consonants from 4 vowels and 5 consonants will be in
and the number of way in which we can arrange the 5 letters is 5!
so, 60 x 5! = 60 x 120 = 7200
and the number of way in which we can arrange the 5 letters is 5!
so, 60 x 5! = 60 x 120 = 7200
Even numbers are formed with three digits such that if 5 is one of the digit then 7 is the next digit. The number of such numbers is (repetition is allowed).- a)360
- b)365
- c)325
- d)none
Correct answer is option 'B'. Can you explain this answer?
Even numbers are formed with three digits such that if 5 is one of the digit then 7 is the next digit. The number of such numbers is (repetition is allowed).
a)
360
b)
365
c)
325
d)
none
|
|
Chirag Verma answered |
5 will be followed by 7, but 7 can come independently, (none is correct)
The number of 6-digit numbers in which the sum of the digits is divisible by 5 is- a)180000
- b)540000
- c)5 x 105
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
The number of 6-digit numbers in which the sum of the digits is divisible by 5 is
a)
180000
b)
540000
c)
5 x 105
d)
none of these
|
|
Chirag Verma answered |
x x x x x x
Wavs 9 10 10 10 10 6
0 cannot be filled in the first place. In other four places any digit can be filled. After filling the first five places, the last place can be filled by 0 or 5, 1 or 6, 2 or 7, 3 or 8, 4 or 9 depending upon whether the sum or five digits filled is of the form 5m, 5m + 4, 5m + 3, 5m + 2 or 5m + 1 respectively.
∴ The required number of numbers = 9 x 104 x 2
Wavs 9 10 10 10 10 6
0 cannot be filled in the first place. In other four places any digit can be filled. After filling the first five places, the last place can be filled by 0 or 5, 1 or 6, 2 or 7, 3 or 8, 4 or 9 depending upon whether the sum or five digits filled is of the form 5m, 5m + 4, 5m + 3, 5m + 2 or 5m + 1 respectively.
∴ The required number of numbers = 9 x 104 x 2
The number of ways in which a score of a 11 can be made from a throw by three persons, each throwing a single die once is :- a)45
- b)18
- c)27
- d)none
Correct answer is option 'C'. Can you explain this answer?
The number of ways in which a score of a 11 can be made from a throw by three persons, each throwing a single die once is :
a)
45
b)
18
c)
27
d)
none
|
|
Chirag Verma answered |
Each value of roll should lie between 1 and 6 and sum should be 11.
= 3 + 4 + 5 + 6 + 5 + 4 = 27 ways
= 3 + 4 + 5 + 6 + 5 + 4 = 27 ways
The number of 5-digit numbers that can be made using the digits 1 and 2 and in which at least one digit is different, is- a)30
- b)31
- c)32
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
The number of 5-digit numbers that can be made using the digits 1 and 2 and in which at least one digit is different, is
a)
30
b)
31
c)
32
d)
none of these
|
Lavanya Patel answered |
Number of 5-digit numbers that can be made using the digits 1 and 2 is a permutation problem. We need to count the number of arrangements of the digits 1 and 2 in a 5-digit number.
1. Counting the total number of 5-digit numbers:
Since we have 2 choices for each digit (1 or 2), the total number of 5-digit numbers that can be formed is given by 2 * 2 * 2 * 2 * 2 = 32. This means there are 32 possible 5-digit numbers using only the digits 1 and 2.
2. Counting the number of 5-digit numbers where all digits are the same:
If all the digits are the same, there can only be two possibilities: 11111 or 22222. So, there are only 2 such numbers.
3. Counting the number of 5-digit numbers where at least one digit is different:
To find the number of 5-digit numbers where at least one digit is different, we need to subtract the number of numbers where all digits are the same from the total number of 5-digit numbers.
Number of 5-digit numbers where at least one digit is different = Total number of 5-digit numbers - Number of 5-digit numbers where all digits are the same
= 32 - 2
= 30
Therefore, the correct answer is option 'A', 30.
1. Counting the total number of 5-digit numbers:
Since we have 2 choices for each digit (1 or 2), the total number of 5-digit numbers that can be formed is given by 2 * 2 * 2 * 2 * 2 = 32. This means there are 32 possible 5-digit numbers using only the digits 1 and 2.
2. Counting the number of 5-digit numbers where all digits are the same:
If all the digits are the same, there can only be two possibilities: 11111 or 22222. So, there are only 2 such numbers.
3. Counting the number of 5-digit numbers where at least one digit is different:
To find the number of 5-digit numbers where at least one digit is different, we need to subtract the number of numbers where all digits are the same from the total number of 5-digit numbers.
Number of 5-digit numbers where at least one digit is different = Total number of 5-digit numbers - Number of 5-digit numbers where all digits are the same
= 32 - 2
= 30
Therefore, the correct answer is option 'A', 30.
The total number of 7 digits numbers the sum of whose digits is even is :- a)9000000
- b)4500000
- c)8100000
- d)none
Correct answer is option 'B'. Can you explain this answer?
The total number of 7 digits numbers the sum of whose digits is even is :
a)
9000000
b)
4500000
c)
8100000
d)
none
|
|
Chirag Verma answered |
Total 7 digit no. = 9x106
sum will be even in half cases
9000000/2 = 4500000
sum will be even in half cases
9000000/2 = 4500000
The number of ways in which we can select four numbers from 1 to 30 so as to exclude every selection of four consecutive numbers is :- a)27399
- b)27378
- c)27405
- d)none
Correct answer is option 'B'. Can you explain this answer?
The number of ways in which we can select four numbers from 1 to 30 so as to exclude every selection of four consecutive numbers is :
a)
27399
b)
27378
c)
27405
d)
none
|
|
Veda Institute answered |
We select from 1 to 30 so as to exclude every selection of four consecutive number in 30C4 - 27 ways
There are 3 candidate for a Mathematics, 5 for chemistry and 4 for a Physics scholarship. In how many ways can the scholarship be awarded.- a)12
- b)60
- c)20
- d)none
Correct answer is option 'B'. Can you explain this answer?
There are 3 candidate for a Mathematics, 5 for chemistry and 4 for a Physics scholarship. In how many ways can the scholarship be awarded.
a)
12
b)
60
c)
20
d)
none
|
|
Chirag Verma answered |
There are 3 candidates of math.
There arc 5 candidates of chemistry.
There arc 4 candidates of physics,
so total way = 3 * 4 * 5 = 60
There arc 5 candidates of chemistry.
There arc 4 candidates of physics,
so total way = 3 * 4 * 5 = 60
A bag contains 3 black, 4 white and 2 red balls, all the balls being different. The number of selections of at most 6 balls cotaining balls of all the colours is- a)42(4!)
- b)26 x 4!
- c)(25-1)(4!)
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
A bag contains 3 black, 4 white and 2 red balls, all the balls being different. The number of selections of at most 6 balls cotaining balls of all the colours is
a)
42(4!)
b)
26 x 4!
c)
(25-1)(4!)
d)
none of these
|
|
Chirag Verma answered |
The required number of selections.
To fill 12 vacancies there are 25 candidates of which 5 are from scheduled casts. If 3 of the vacancies are reserved for scheduled casts candidates while the rest are open to all. The number of ways in which the selections can be made is :- a)5C3 * 22C9
- b)4973200
- c)27C12
- d)none
Correct answer is option 'A'. Can you explain this answer?
To fill 12 vacancies there are 25 candidates of which 5 are from scheduled casts. If 3 of the vacancies are reserved for scheduled casts candidates while the rest are open to all. The number of ways in which the selections can be made is :
a)
5C3 * 22C9
b)
4973200
c)
27C12
d)
none
|
|
Veda Institute answered |
Total vacancies = 12
total candidate = 25
number of scheduled caste = 5 out of 25.
reserved vacancies = 3
total ways of selection = 5C3 x 22C9
total candidate = 25
number of scheduled caste = 5 out of 25.
reserved vacancies = 3
total ways of selection = 5C3 x 22C9
A businessman hosts a dinner to 21 guests. He is having 2 round tables which can accommondate 15 and 6 persons. How many ways are there to host dinner?- a)21C6 * 120 * 14!
- b)21C15 * 14!
- c)21C6 *15! * 6!
- d)none
Correct answer is option 'D'. Can you explain this answer?
A businessman hosts a dinner to 21 guests. He is having 2 round tables which can accommondate 15 and 6 persons. How many ways are there to host dinner?
a)
21C6 * 120 * 14!
b)
21C15 * 14!
c)
21C6 *15! * 6!
d)
none
|
|
Veda Institute answered |
Total guest = 21, guests are
seated in round table respectively are 15 and 6
Total ways = 21C15 x 14!5!
seated in round table respectively are 15 and 6
Total ways = 21C15 x 14!5!
The number of 6 digit numbers that can be made with the digits 0. 1,2, 3, 4 and 5 so that even digits occupy odd places, is- a)24
- b)36
- c)48
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
The number of 6 digit numbers that can be made with the digits 0. 1,2, 3, 4 and 5 so that even digits occupy odd places, is
a)
24
b)
36
c)
48
d)
none of these
|
|
Veda Institute answered |
x | x | x | Crosses can be filled in 3P3 - 2P2, ways
(∴ 0 cannot go in the first place from the left).
The remaining places can be filled in 3! ways.
∴ the required number of numbers = (3P3-2P2) x 3!
(∴ 0 cannot go in the first place from the left).
The remaining places can be filled in 3! ways.
∴ the required number of numbers = (3P3-2P2) x 3!
There are 20 questions in a question paper. If no two students solve the same combination of questions but solve equal number of questions then the maximum number of students who appeared in the examination is- a)20C9
- b)20C11
- c)20C10
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
There are 20 questions in a question paper. If no two students solve the same combination of questions but solve equal number of questions then the maximum number of students who appeared in the examination is
a)
20C9
b)
20C11
c)
20C10
d)
none of these
|
|
Chirag Verma answered |
If r questions are solved by each student then the number of possible selections of questions is 20Cr
∴ the number of students =20Cr
(∴ each student has solved different combination of questions)
∴ the maximum number of students = maximum value of 20Cr = 20C10, because 20C10 is the largest among 20C0, 20C1, ....20C20 - being the middle one.
∴ the number of students =20Cr
(∴ each student has solved different combination of questions)
∴ the maximum number of students = maximum value of 20Cr = 20C10, because 20C10 is the largest among 20C0, 20C1, ....20C20 - being the middle one.
The number of arrangements of the letters of the word BHARAT taking 3 at a time is- a)72
- b)120
- c)14
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
The number of arrangements of the letters of the word BHARAT taking 3 at a time is
a)
72
b)
120
c)
14
d)
none of these
|
Gauri Desai answered |
To find the number of arrangements of the letters of the word BHARAT taking 3 at a time, we can use the concept of permutations.
Permutations refer to the arrangement of objects in a specific order. In this case, we need to find the number of arrangements of the letters of the word BHARAT taking 3 at a time, which means we need to arrange 3 letters out of the total 6 letters of the word.
Using the formula for permutations, we can calculate the number of arrangements as follows:
nPr = n! / (n-r)!
Where n represents the total number of objects and r represents the number of objects being arranged.
In this case, n = 6 (total number of letters in the word BHARAT) and r = 3 (number of letters being arranged).
Plugging in these values into the formula, we get:
6P3 = 6! / (6-3)!
= 6! / 3!
= (6 * 5 * 4 * 3 * 2 * 1) / (3 * 2 * 1)
= 6 * 5 * 4
= 120
Therefore, the number of arrangements of the letters of the word BHARAT taking 3 at a time is 120.
Hence, the correct answer is option 'B' - 120.
Permutations refer to the arrangement of objects in a specific order. In this case, we need to find the number of arrangements of the letters of the word BHARAT taking 3 at a time, which means we need to arrange 3 letters out of the total 6 letters of the word.
Using the formula for permutations, we can calculate the number of arrangements as follows:
nPr = n! / (n-r)!
Where n represents the total number of objects and r represents the number of objects being arranged.
In this case, n = 6 (total number of letters in the word BHARAT) and r = 3 (number of letters being arranged).
Plugging in these values into the formula, we get:
6P3 = 6! / (6-3)!
= 6! / 3!
= (6 * 5 * 4 * 3 * 2 * 1) / (3 * 2 * 1)
= 6 * 5 * 4
= 120
Therefore, the number of arrangements of the letters of the word BHARAT taking 3 at a time is 120.
Hence, the correct answer is option 'B' - 120.
Chapter doubts & questions for Permutations and Combinations - Topic-wise Tests & Solved Examples for Mathematics 2025 is part of Mathematics exam preparation. The chapters have been prepared according to the Mathematics exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Mathematics 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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