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 :

^{20}C_{10}.2^{10}
^{20}C_{10} 3^{10}
^{20}C_{10}.3^{10}
^{20}C_{10}.2^{20}

^{20}C_{10}.2^{10}^{20}C_{10}3^{10}^{20}C_{10}.3^{10}^{20}C_{10}.2^{20}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: ^{20}C_{10}.2^{10}

### 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 ^{9}P_{5} + 5 ^{9}P_{4} = , 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 ^{9}P_{5} + 5 ^{9}P_{4} = , 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) :

maths-General

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

Maths-

### If x is real, then maximum value of is

### If x is real, then maximum value of is

Maths-General

Maths-

### The value of 'c' of Lagrange's mean value theorem for is

### The value of 'c' of Lagrange's mean value theorem for is

Maths-General

Maths-

### The value of 'c' of Rolle's mean value theorem for is

### The value of 'c' of Rolle's mean value theorem for is

Maths-General

Maths-

### The value of 'c' of Rolle's theorem for – on [–1, 1] is

### The value of 'c' of Rolle's theorem for – on [–1, 1] is

Maths-General

Maths-

### For in [5, 7]

### For in [5, 7]

Maths-General

Maths-

### The value of 'c' in Lagrange's mean value theorem for in [0, 1] is

### The value of 'c' in Lagrange's mean value theorem for in [0, 1] is

Maths-General

Maths-

### The value of 'c' in Lagrange's mean value theorem for in [0, 2] is

### The value of 'c' in Lagrange's mean value theorem for in [0, 2] is

Maths-General

Maths-

### The equation represents

### The equation represents

Maths-General

Maths-

### 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

Maths-General

Maths-

### The radius of the circle is

### The radius of the circle is

Maths-General