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
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) :
Alternatively, we can use the formula for the sum of numbers as
We can also solve this problem by writing all the possible numbers and finding the sum of them which will be time taking and make us confused. We should know that the value of the digits is determined by the place where they were present. We should check whether there is zero in the given digits and whether there are any repetitions present in the numbers. Similarly, we can expect problems to find the sum of numbers formed by these digits with repetition allowed.
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) :
Alternatively, we can use the formula for the sum of numbers as
We can also solve this problem by writing all the possible numbers and finding the sum of them which will be time taking and make us confused. We should know that the value of the digits is determined by the place where they were present. We should check whether there is zero in the given digits and whether there are any repetitions present in the numbers. Similarly, we can expect problems to find the sum of numbers formed by these digits with repetition allowed.
Total number of divisors of 480, that are of the form 4n + 2, n 
0, is equal to :
We can also solve this question by writing 4n + 2 = 2(2n + 1) where 2n + 1 is always an odd number. So, when all odd divisors will be multiplied by 2, we will get the divisors that we require. Hence, we can say a number of divisors of 4n + 2 form is the same as the number of odd divisors for 480.
Total number of divisors of 480, that are of the form 4n + 2, n 0, is equal to :
We can also solve this question by writing 4n + 2 = 2(2n + 1) where 2n + 1 is always an odd number. So, when all odd divisors will be multiplied by 2, we will get the divisors that we require. Hence, we can say a number of divisors of 4n + 2 form is the same as the number of odd divisors for 480.
If 9P5 + 5 9P4 = 
, then r =
If 9P5 + 5 9P4 = , then r =
Assertion (A) :If 
, then 
Reason (R) :
Assertion (A) :If , then
Reason (R) :
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
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
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in [0, 1] is
in [0, 2] is
represents
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