Maths-
General
Easy
Question
A person predicts the outcome of 20 cricket matches of his home team. Each match can result either in a win, loss or tie for the home team. Total number of ways in which he can make the predictions so that exactly 10 predictions are correct, is equal to :
- 20C10.210
- 20C10 310
- 20C10.310
- 20C10.220
Hint:
There are total 20 matches and the outcome can either be win, lose or tie. We have to find the number of ways in which exactly 10 predictions are correct which can be shown by
The correct answer is: 20C10.210
There are total 20 matches and the outcome can either be win, lose or tie.
We have to find the number of ways in which exactly 10 predictions are correct which can be shown by ways in which his prediction is correct.
And in the remaining 10 matches, he makes wrong predictions i.e. out of 3 outcomes (win, lose, tie) he can pick 2 outcomes out of 3 , which can be done in 
ways.
Thus, total number of ways in which he can make the predictions so that exactly 10 predictions are correct, is equal to 
Related Questions to study
Maths-
The sum of all the numbers that can be formed with the digits 2, 3, 4, 5 taken all at a time is (repetition is not allowed) :
Detailed Solution
According to the problem, we need to find the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time.
Let us first fix a number in a unit place and find the total number of words possible due on fixing this number.
We need to arrange the remaining three places with three digits. We know that the number of ways of arranging n objects in n places is n! ways.
So, we get 3! = 6 numbers on fixing the unit place with a particular digit.
Now, let us find the sum of all digits.
We use the same digits in ten, hundred and thousand places also. So, the sum of those digits will also be 84 but with the multiplication of its place value.
i.e., We multiply the sum of the digits in tenth place with 10, hundredth place with 100 and so on. We then add these sums.
Thus, the sum of all the numbers that can be formed with the digits 2, 3, 4, 5 taken all at a time is (repetition is not allowed) : 93324
According to the problem, we need to find the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time.
Let us first fix a number in a unit place and find the total number of words possible due on fixing this number.
We need to arrange the remaining three places with three digits. We know that the number of ways of arranging n objects in n places is n! ways.
So, we get 3! = 6 numbers on fixing the unit place with a particular digit.
Now, let us find the sum of all digits.
We use the same digits in ten, hundred and thousand places also. So, the sum of those digits will also be 84 but with the multiplication of its place value.
i.e., We multiply the sum of the digits in tenth place with 10, hundredth place with 100 and so on. We then add these sums.
Thus, the sum of all the numbers that can be formed with the digits 2, 3, 4, 5 taken all at a time is (repetition is not allowed) : 93324
The sum of all the numbers that can be formed with the digits 2, 3, 4, 5 taken all at a time is (repetition is not allowed) :
Maths-General
Detailed Solution
According to the problem, we need to find the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time.
Let us first fix a number in a unit place and find the total number of words possible due on fixing this number.
We need to arrange the remaining three places with three digits. We know that the number of ways of arranging n objects in n places is n! ways.
So, we get 3! = 6 numbers on fixing the unit place with a particular digit.
Now, let us find the sum of all digits.
We use the same digits in ten, hundred and thousand places also. So, the sum of those digits will also be 84 but with the multiplication of its place value.
i.e., We multiply the sum of the digits in tenth place with 10, hundredth place with 100 and so on. We then add these sums.
Thus, the sum of all the numbers that can be formed with the digits 2, 3, 4, 5 taken all at a time is (repetition is not allowed) : 93324
According to the problem, we need to find the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time.
Let us first fix a number in a unit place and find the total number of words possible due on fixing this number.
We need to arrange the remaining three places with three digits. We know that the number of ways of arranging n objects in n places is n! ways.
So, we get 3! = 6 numbers on fixing the unit place with a particular digit.
Now, let us find the sum of all digits.
We use the same digits in ten, hundred and thousand places also. So, the sum of those digits will also be 84 but with the multiplication of its place value.
i.e., We multiply the sum of the digits in tenth place with 10, hundredth place with 100 and so on. We then add these sums.
Thus, the sum of all the numbers that can be formed with the digits 2, 3, 4, 5 taken all at a time is (repetition is not allowed) : 93324
Maths-
Total number of divisors of 480, that are of the form 4n + 2, n 
0, is equal to :
Detailed Solution
In this question, we have been asked to find the total number of divisors of 480 which are of the form 4n + 2, n 0
In order to solve this question, we should know that the number of the divisor of any number
where a, b, c are prime numbers and is given by (m + 1) (n + 1) (p + 1)…..
We know that 480 can be expressed as
Now, we have been asked to find the number of divisors which are of the form 4n + 2 = 2 (2n + 1), which means odd divisors cannot be a part of the solution.
Now, we can say the total number of even divisors are = all divisors – odd divisor = 24 - 4 = 20
Now, we have been given that the divisor should be of 4n + 2, which means they should not be a multiple of 4 but multiple of 2. For that, we will subtract the multiple of 4 which are divisor of 480 from the even divisors.
And, we know that,
Hence, we can say that there are 16 divisors of 480 which are multiple of 4.
Thus, total number of divisors of 480, that are of the form 4n + 2, n 0, is equal to 4.
In this question, we have been asked to find the total number of divisors of 480 which are of the form 4n + 2, n
0 In order to solve this question, we should know that the number of the divisor of any number
where a, b, c are prime numbers and is given by (m + 1) (n + 1) (p + 1)…..We know that 480 can be expressed as
Now, we have been asked to find the number of divisors which are of the form 4n + 2 = 2 (2n + 1), which means odd divisors cannot be a part of the solution.
Now, we can say the total number of even divisors are = all divisors – odd divisor = 24 - 4 = 20
Now, we have been given that the divisor should be of 4n + 2, which means they should not be a multiple of 4 but multiple of 2. For that, we will subtract the multiple of 4 which are divisor of 480 from the even divisors.
And, we know that,
Hence, we can say that there are 16 divisors of 480 which are multiple of 4.
Thus, total number of divisors of 480, that are of the form 4n + 2, n
0, is equal to 4.Total number of divisors of 480, that are of the form 4n + 2, n 0, is equal to :
Maths-General
Detailed Solution
In this question, we have been asked to find the total number of divisors of 480 which are of the form 4n + 2, n 0
In order to solve this question, we should know that the number of the divisor of any number
where a, b, c are prime numbers and is given by (m + 1) (n + 1) (p + 1)…..
We know that 480 can be expressed as
Now, we have been asked to find the number of divisors which are of the form 4n + 2 = 2 (2n + 1), which means odd divisors cannot be a part of the solution.
Now, we can say the total number of even divisors are = all divisors – odd divisor = 24 - 4 = 20
Now, we have been given that the divisor should be of 4n + 2, which means they should not be a multiple of 4 but multiple of 2. For that, we will subtract the multiple of 4 which are divisor of 480 from the even divisors.
And, we know that,
Hence, we can say that there are 16 divisors of 480 which are multiple of 4.
Thus, total number of divisors of 480, that are of the form 4n + 2, n 0, is equal to 4.
In this question, we have been asked to find the total number of divisors of 480 which are of the form 4n + 2, n
0 In order to solve this question, we should know that the number of the divisor of any number
where a, b, c are prime numbers and is given by (m + 1) (n + 1) (p + 1)…..We know that 480 can be expressed as
Now, we have been asked to find the number of divisors which are of the form 4n + 2 = 2 (2n + 1), which means odd divisors cannot be a part of the solution.
Now, we can say the total number of even divisors are = all divisors – odd divisor = 24 - 4 = 20
Now, we have been given that the divisor should be of 4n + 2, which means they should not be a multiple of 4 but multiple of 2. For that, we will subtract the multiple of 4 which are divisor of 480 from the even divisors.
And, we know that,
Hence, we can say that there are 16 divisors of 480 which are multiple of 4.
Thus, total number of divisors of 480, that are of the form 4n + 2, n
0, is equal to 4.Maths-
If 9P5 + 5 9P4 = 
, then r =
Detailed Solution
Here we need to find the value of the given variable.
The given expression is :
Here we need to find the value of the given variable.
The given expression is :
If 9P5 + 5 9P4 = , then r =
Maths-General
Detailed Solution
Here we need to find the value of the given variable.
The given expression is :
Here we need to find the value of the given variable.
The given expression is :
maths-
Assertion (A) :If 
, then 
Reason (R) :
Assertion (A) :If , then
Reason (R) :
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physics-
A particle is released from a height. At certain height its kinetic energy is three times its potential energy. The height and speed of the particle at that instant are respectively
Velocity at when dropped from where
Or (i)
Potential energy at (ii)
Kinetic energy potential energy
Or (i)
Potential energy at (ii)
Kinetic energy potential energy
A particle is released from a height. At certain height its kinetic energy is three times its potential energy. The height and speed of the particle at that instant are respectively
physics-General
Velocity at when dropped from where
Or (i)
Potential energy at (ii)
Kinetic energy potential energy
Or (i)
Potential energy at (ii)
Kinetic energy potential energy
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If x is real, then maximum value of 
is
If x is real, then maximum value of is
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is
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is
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– 
on [–1, 1] is
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in [5, 7]
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in [0, 2] is
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The equation 
represents
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The polar equation of the circle whose end points of the diameter are 
and 
is
The polar equation of the circle whose end points of the diameter are and is
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The radius of the circle 
is
The radius of the circle is
Maths-General
