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 :

  1. 20C10.210    
  2. 20C10 310
       
  3. 20C10.310    
  4. 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 byC presuperscript 20 subscript 10

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 C presuperscript 20 subscript 10 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 2 to the power of 10 ways.
    Thus, total number of ways in which he can make the predictions so that exactly 10 predictions are correct, is equal to C presuperscript 20 subscript 10 cross times 2 to the power of 10

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    Related Questions to study

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    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.
    W e space g e t space s u m space a s space 2 space plus space 3 space plus space 4 space plus space 5 space equals space 14
    N o w comma space w e space g e t space a space s u m space o f space d i g i t s space i n space u n i t s space p l a c e space f o r space a l l space t h e space n u m b e r s space a s space 14 cross times 6 equals space 84.
    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.
    S o comma space w e space g e t space t h e space s u m space o f space a l l space t h e space n u m b e r s space t h a t space c a n space b e space f o r m e d space w i t h space t h e space d i g i t s space 2 comma 3 comma 4 comma 5 space t a k e n space a l l space a t space a space t i m e space i s space
left parenthesis 84 cross times 1000 right parenthesis plus left parenthesis 84 cross times 100 right parenthesis plus left parenthesis 84 cross times 10 right parenthesis plus left parenthesis 84 cross times 1 right parenthesis space equals space 93324
    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.
    W e space g e t space s u m space a s space 2 space plus space 3 space plus space 4 space plus space 5 space equals space 14
    N o w comma space w e space g e t space a space s u m space o f space d i g i t s space i n space u n i t s space p l a c e space f o r space a l l space t h e space n u m b e r s space a s space 14 cross times 6 equals space 84.
    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.
    S o comma space w e space g e t space t h e space s u m space o f space a l l space t h e space n u m b e r s space t h a t space c a n space b e space f o r m e d space w i t h space t h e space d i g i t s space 2 comma 3 comma 4 comma 5 space t a k e n space a l l space a t space a space t i m e space i s space
left parenthesis 84 cross times 1000 right parenthesis plus left parenthesis 84 cross times 100 right parenthesis plus left parenthesis 84 cross times 10 right parenthesis plus left parenthesis 84 cross times 1 right parenthesis space equals space 93324
    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
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    Total number of divisors of 480, that are of the form 4n + 2, n greater or equal than 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 greater or equal than 0 
    In order to solve this question, we should know that the number of the divisor of any number
    x equals a to the power of m space end exponent b to the power of n c to the power of p.... space space 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 480 space equals space 2 to the power of 5.3.5
    S o comma space a c c o r d i n g space t o space t h e space f o r m u l a comma space t h e space t o t a l space n u m b e r space o f space d i v i s o r s space o f space 480 space a r e
space left parenthesis 5 space plus space 1 right parenthesis space left parenthesis 1 space plus space 1 right parenthesis space left parenthesis 1 space plus space 1 right parenthesis space equals space 6 cross times 2 cross times 2 equals 24 space.
    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. 
    S o comma space t h e space t o t a l space n u m b e r space o f space o d d space d i v i s o r s space t h a t space a r e space p o s s i b l e space a r e space left parenthesis 1 space plus space 1 right parenthesis space left parenthesis 1 space plus space 1 right parenthesis space equals space 2 cross times 2 equals 4 space comma space a c c o r d i n g space t o space t h e space p r o p e r t y.
    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, 480 space equals space 2 to the power of 5.3.5 
    space S o comma space t h e space n u m b e r space o f space d i v i s o r s space t h a t space a r e space m u l t i p l e s space o f space 4 space a r e space left parenthesis 3 space plus space 1 right parenthesis space left parenthesis 1 space plus space 1 right parenthesis space left parenthesis 1 space plus space 1 right parenthesis space equals space 4 cross times 2 cross times 2 space space equals space 16
    Hence, we can say that there are 16 divisors of 480 which are multiple of 4.
    S o comma space t h e space t o t a l space n u m b e r space o f space d i v i s o r s space w h i c h space a r e space e v e n space b u t space n o t space d i v i s i b l e space b y space 2 space c a n space b e space g i v e n space b y space 20 space – space 16 space equals space 4.
    Thus, total number of divisors of 480, that are of the form 4n + 2, n greater or equal than 0, is equal to 4.

    Total number of divisors of 480, that are of the form 4n + 2, n greater or equal than 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 greater or equal than 0 
    In order to solve this question, we should know that the number of the divisor of any number
    x equals a to the power of m space end exponent b to the power of n c to the power of p.... space space 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 480 space equals space 2 to the power of 5.3.5
    S o comma space a c c o r d i n g space t o space t h e space f o r m u l a comma space t h e space t o t a l space n u m b e r space o f space d i v i s o r s space o f space 480 space a r e
space left parenthesis 5 space plus space 1 right parenthesis space left parenthesis 1 space plus space 1 right parenthesis space left parenthesis 1 space plus space 1 right parenthesis space equals space 6 cross times 2 cross times 2 equals 24 space.
    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. 
    S o comma space t h e space t o t a l space n u m b e r space o f space o d d space d i v i s o r s space t h a t space a r e space p o s s i b l e space a r e space left parenthesis 1 space plus space 1 right parenthesis space left parenthesis 1 space plus space 1 right parenthesis space equals space 2 cross times 2 equals 4 space comma space a c c o r d i n g space t o space t h e space p r o p e r t y.
    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, 480 space equals space 2 to the power of 5.3.5 
    space S o comma space t h e space n u m b e r space o f space d i v i s o r s space t h a t space a r e space m u l t i p l e s space o f space 4 space a r e space left parenthesis 3 space plus space 1 right parenthesis space left parenthesis 1 space plus space 1 right parenthesis space left parenthesis 1 space plus space 1 right parenthesis space equals space 4 cross times 2 cross times 2 space space equals space 16
    Hence, we can say that there are 16 divisors of 480 which are multiple of 4.
    S o comma space t h e space t o t a l space n u m b e r space o f space d i v i s o r s space w h i c h space a r e space e v e n space b u t space n o t space d i v i s i b l e space b y space 2 space c a n space b e space g i v e n space b y space 20 space – space 16 space equals space 4.
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    If 9P5 + 5 9P4 = scriptbase P subscript r end scriptbase presuperscript 10, then r =

    Detailed Solution
     Here we need to find the value of the given variable.
    The given expression is :
    scriptbase P subscript 5 space plus space 5. end scriptbase presuperscript 9 scriptbase P subscript 4 space equals space scriptbase P subscript r end scriptbase presuperscript 10 end scriptbase presuperscript 9

     
N o w comma space w e space w i l l space u s e space t h i s space f o r m u l a space o f space p e r m u t a t i o n space
scriptbase P subscript r space equals space fraction numerator n factorial over denominator left parenthesis n minus r right parenthesis factorial end fraction end scriptbase presuperscript n comma space i n space t h e space a b o v e space e q u a t i o n

T h e r e f o r e comma space w e space g e t space

rightwards double arrow fraction numerator 9 factorial over denominator left parenthesis 9 minus 5 right parenthesis factorial end fraction space plus space 5. space fraction numerator 9 factorial over denominator left parenthesis 9 minus 4 right parenthesis factorial end fraction space equals space fraction numerator 10 factorial over denominator left parenthesis 10 minus r right parenthesis factorial end fraction space space space space

rightwards double arrow fraction numerator 9 factorial over denominator left parenthesis 4 right parenthesis factorial end fraction space plus space 5. space fraction numerator 9 factorial over denominator left parenthesis 5 right parenthesis factorial end fraction space equals space fraction numerator 10 factorial over denominator left parenthesis 10 minus r right parenthesis factorial end fraction

D i v i d i n g space b o t h space s i d e s space b y space 9 factorial space w e space g e t


rightwards double arrow fraction numerator 1 over denominator left parenthesis 4 right parenthesis factorial end fraction space plus space 5. space fraction numerator 1 over denominator left parenthesis 5 right parenthesis factorial end fraction space equals space fraction numerator 10 over denominator left parenthesis 9 minus r right parenthesis factorial end fraction............. space space space space space space space space space open parentheses fraction numerator 10 factorial over denominator 9 factorial end fraction space equals space fraction numerator 10 space cross times 9 factorial over denominator 9 factorial end fraction space equals space 10 close parentheses

rightwards double arrow fraction numerator 1 over denominator left parenthesis 4 right parenthesis factorial end fraction space plus space 5. space fraction numerator 1 over denominator 5 space cross times 4 factorial end fraction space equals space fraction numerator 10 over denominator left parenthesis 9 minus r right parenthesis factorial end fraction

rightwards double arrow fraction numerator 1 over denominator left parenthesis 4 right parenthesis factorial end fraction space plus space space fraction numerator 1 over denominator 4 factorial end fraction space equals space fraction numerator 10 over denominator left parenthesis 9 minus r right parenthesis factorial end fraction

rightwards double arrow fraction numerator 2 over denominator 4 factorial end fraction space equals space fraction numerator 10 over denominator left parenthesis 9 minus r right parenthesis factorial end fraction

D i v i d i n g space b y space 2 space o n space b o t h space s i d e s

rightwards double arrow fraction numerator 1 over denominator 4 factorial end fraction space equals space fraction numerator 5 over denominator left parenthesis 9 minus r right parenthesis factorial end fraction

O n space c r o s s space m u l t i p l y i n g

rightwards double arrow left parenthesis 9 minus r right parenthesis factorial space equals space 5 space cross times 4 factorial

rightwards double arrow left parenthesis 9 minus r right parenthesis factorial space equals space 5 factorial

O n space e q u a t i n g space t h e space t w o space f a c t o r i a l s comma space w e space g e t

rightwards double arrow 9 space minus space r space equals space 5
rightwards double arrow r space equals space 4

    If 9P5 + 5 9P4 = scriptbase P subscript r end scriptbase presuperscript 10, then r =

    Maths-General
    Detailed Solution
     Here we need to find the value of the given variable.
    The given expression is :
    scriptbase P subscript 5 space plus space 5. end scriptbase presuperscript 9 scriptbase P subscript 4 space equals space scriptbase P subscript r end scriptbase presuperscript 10 end scriptbase presuperscript 9

     
N o w comma space w e space w i l l space u s e space t h i s space f o r m u l a space o f space p e r m u t a t i o n space
scriptbase P subscript r space equals space fraction numerator n factorial over denominator left parenthesis n minus r right parenthesis factorial end fraction end scriptbase presuperscript n comma space i n space t h e space a b o v e space e q u a t i o n

T h e r e f o r e comma space w e space g e t space

rightwards double arrow fraction numerator 9 factorial over denominator left parenthesis 9 minus 5 right parenthesis factorial end fraction space plus space 5. space fraction numerator 9 factorial over denominator left parenthesis 9 minus 4 right parenthesis factorial end fraction space equals space fraction numerator 10 factorial over denominator left parenthesis 10 minus r right parenthesis factorial end fraction space space space space

rightwards double arrow fraction numerator 9 factorial over denominator left parenthesis 4 right parenthesis factorial end fraction space plus space 5. space fraction numerator 9 factorial over denominator left parenthesis 5 right parenthesis factorial end fraction space equals space fraction numerator 10 factorial over denominator left parenthesis 10 minus r right parenthesis factorial end fraction

D i v i d i n g space b o t h space s i d e s space b y space 9 factorial space w e space g e t


rightwards double arrow fraction numerator 1 over denominator left parenthesis 4 right parenthesis factorial end fraction space plus space 5. space fraction numerator 1 over denominator left parenthesis 5 right parenthesis factorial end fraction space equals space fraction numerator 10 over denominator left parenthesis 9 minus r right parenthesis factorial end fraction............. space space space space space space space space space open parentheses fraction numerator 10 factorial over denominator 9 factorial end fraction space equals space fraction numerator 10 space cross times 9 factorial over denominator 9 factorial end fraction space equals space 10 close parentheses

rightwards double arrow fraction numerator 1 over denominator left parenthesis 4 right parenthesis factorial end fraction space plus space 5. space fraction numerator 1 over denominator 5 space cross times 4 factorial end fraction space equals space fraction numerator 10 over denominator left parenthesis 9 minus r right parenthesis factorial end fraction

rightwards double arrow fraction numerator 1 over denominator left parenthesis 4 right parenthesis factorial end fraction space plus space space fraction numerator 1 over denominator 4 factorial end fraction space equals space fraction numerator 10 over denominator left parenthesis 9 minus r right parenthesis factorial end fraction

rightwards double arrow fraction numerator 2 over denominator 4 factorial end fraction space equals space fraction numerator 10 over denominator left parenthesis 9 minus r right parenthesis factorial end fraction

D i v i d i n g space b y space 2 space o n space b o t h space s i d e s

rightwards double arrow fraction numerator 1 over denominator 4 factorial end fraction space equals space fraction numerator 5 over denominator left parenthesis 9 minus r right parenthesis factorial end fraction

O n space c r o s s space m u l t i p l y i n g

rightwards double arrow left parenthesis 9 minus r right parenthesis factorial space equals space 5 space cross times 4 factorial

rightwards double arrow left parenthesis 9 minus r right parenthesis factorial space equals space 5 factorial

O n space e q u a t i n g space t h e space t w o space f a c t o r i a l s comma space w e space g e t

rightwards double arrow 9 space minus space r space equals space 5
rightwards double arrow r space equals space 4
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    Assertion (A) :If fraction numerator 5 x plus 1 over denominator left parenthesis x plus 2 right parenthesis left parenthesis x minus 1 right parenthesis end fraction equals fraction numerator A over denominator x plus 2 end fraction plus fraction numerator B over denominator x minus 1 end fraction, then A equals 3 comma B equals 2
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