A person predicts the outcome of 20 cricket matches of his home-Turito

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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
²⁰C₁₀.2¹⁰
²⁰C₁₀.3¹⁰
²⁰C₁₀.2²⁰

Hint:

Total number of ways in which he can make the predictions so that exactly 10 predictions are correct =
$N u m b e r space o f space c o r r e c t space p r e d i c t i o n s space cross times space N u m b e r space o f space i n c o r r e c t space p r e d i c t i o n s$

The correct answer is: ²⁰C₁₀.2¹⁰

Matches whose prediction are correct can be selected in $C presuperscript 20 subscript 10$ ways.
Since each match can result either in a win, loss or tie for the home team.
Now, each wrong prediction can be made in two ways (i.e. the correct result is win and the person predicts either lose or tie)
and there are 10 matches for which he predicted wrong.
Total number of ways in which he can make the predictions so that exactly 10 predictions are correct =
$N u m b e r space o f space c o r r e c t space p r e d i c t i o n s space cross times space N u m b e r space o f space i n c o r r e c t space p r e d i c t i o n s$
$C presuperscript 20 subscript 10 cross times space 2 to the power of 10$
Thus, total number of ways in which he can make the predictions so that exactly 10 predictions are correct, is equal to $C presuperscript 20 subscript 10 cross times space 2 to the power of 10$

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The foot of the perpendicular from the point $left parenthesis 3 , 3 pi divided by 4 right parenthesis$ on the line $r left parenthesis c o s space theta minus s i n space theta right parenthesis equals 6 square root of 2$ is

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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) :

Complete step-by-step answer:

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 get sum as

2 + 3 + 4 + 5 = 14

2 + 3 + 4 + 5 = 14

2 + 3 + 4 + 5 = 14

So, we get the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time is

(84 \times 1000) + (84 \times 100) + (84 \times 10) + (84 \times 1)

(84 \times 1000) + (84 \times 100) + (84 \times 10) + (84 \times 1)

(84 \times 1000) + (84 \times 100) + (84 \times 10) + (84 \times 1)

(84 \times 1000) + (84 \times 100) + (84 \times 10) + (84 \times 1)

(84 \times 1000) + (84 \times 100) + (84 \times 10) + (84 \times 1)

84000 + 8400 + 840 + 84

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) :

Complete step-by-step answer:

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 get sum as

2 + 3 + 4 + 5 = 14

2 + 3 + 4 + 5 = 14

2 + 3 + 4 + 5 = 14

So, we get the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time is

(84 \times 1000) + (84 \times 100) + (84 \times 10) + (84 \times 1)

(84 \times 1000) + (84 \times 100) + (84 \times 10) + (84 \times 1)

(84 \times 1000) + (84 \times 100) + (84 \times 10) + (84 \times 1)

(84 \times 1000) + (84 \times 100) + (84 \times 10) + (84 \times 1)

(84 \times 1000) + (84 \times 100) + (84 \times 10) + (84 \times 1)

84000 + 8400 + 840 + 84

Total number of divisors of 480, that are of the form 4n + 2, n $greater or equal than$ 0, is equal to :

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
To solve this question, we should know that the total number of divisors of any number x of the form

a to the power of m b to the power of n c to the power of p...... space w h e r e space a comma b comma 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

cross times 3 cross times 5

So, according to the formula, the total number of divisors of 480 are (5 + 1) (1 + 1) (1 + 1) =

6 \times 2 \times 2 = 24

6 \times 2 \times 2 = 24

2 \times 2 = 4

, according to the property.
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 squared space left parenthesis space 2 to the power of 8 cross times 3 cross times 5 right parenthesis

So, the number of divisors that are multiples of 4 are (3 + 1) (1 + 1) (1 + 1) =

4 \times 2 \times 2

16.
Hence, we can say that there are 16 divisors of 480 which are multiple of 4.
So, the total number of divisors which are even but not divisible by 2 can be given by 20 – 16 = 4.
Hence, we can say that there are 4 divisors of 480 that are of 4n + 2 form,

n \geq 0

.

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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 :

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
To solve this question, we should know that the total number of divisors of any number x of the form

a to the power of m b to the power of n c to the power of p...... space w h e r e space a comma b comma 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

cross times 3 cross times 5

So, according to the formula, the total number of divisors of 480 are (5 + 1) (1 + 1) (1 + 1) =

6 \times 2 \times 2 = 24

6 \times 2 \times 2 = 24

2 \times 2 = 4

, according to the property.
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 squared space left parenthesis space 2 to the power of 8 cross times 3 cross times 5 right parenthesis

So, the number of divisors that are multiples of 4 are (3 + 1) (1 + 1) (1 + 1) =

4 \times 2 \times 2

16.
Hence, we can say that there are 16 divisors of 480 which are multiple of 4.
So, the total number of divisors which are even but not divisible by 2 can be given by 20 – 16 = 4.
Hence, we can say that there are 4 divisors of 480 that are of 4n + 2 form,

n \geq 0

.

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If ⁹P₅ + 5 ⁹P₄ = ¹⁰P_r, then r =

P presuperscript 9 subscript 5 space plus space 5 space cross times P presuperscript 9 subscript 4 space equals space P presuperscript 10 subscript r

Using Formula :

P presuperscript n subscript r space equals space fraction numerator n factorial over denominator left parenthesis n minus r right parenthesis factorial end fraction

fraction numerator 9 factorial over denominator left parenthesis 9 minus 5 right parenthesis factorial end fraction space plus space 5 cross times 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

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

Dividing both sides by 9!

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

fraction numerator 5 over denominator 5 factorial end fraction plus space fraction numerator 5 over denominator 5 factorial end fraction equals space fraction numerator 10 over denominator left parenthesis 10 minus r right parenthesis factorial end fraction fraction numerator 10 over denominator 5 factorial end fraction space equals space fraction numerator 10 over denominator left parenthesis 10 minus r right parenthesis factorial end fraction E q u a t i n g space t h e space d e n o m i n a t o r s 5 factorial space equals space left parenthesis 10 minus r right parenthesis factorial 5 space equals space 10 space minus space r r space equals space 5

If ⁹P₅ + 5 ⁹P₄ = ¹⁰P_r, then r =

P presuperscript 9 subscript 5 space plus space 5 space cross times P presuperscript 9 subscript 4 space equals space P presuperscript 10 subscript r

Using Formula :

P presuperscript n subscript r space equals space fraction numerator n factorial over denominator left parenthesis n minus r right parenthesis factorial end fraction

fraction numerator 9 factorial over denominator left parenthesis 9 minus 5 right parenthesis factorial end fraction space plus space 5 cross times 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

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

Dividing both sides by 9!

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

fraction numerator 5 over denominator 5 factorial end fraction plus space fraction numerator 5 over denominator 5 factorial end fraction equals space fraction numerator 10 over denominator left parenthesis 10 minus r right parenthesis factorial end fraction fraction numerator 10 over denominator 5 factorial end fraction space equals space fraction numerator 10 over denominator left parenthesis 10 minus r right parenthesis factorial end fraction E q u a t i n g space t h e space d e n o m i n a t o r s 5 factorial space equals space left parenthesis 10 minus r right parenthesis factorial 5 space equals space 10 space minus space r r space equals space 5

The number of proper divisors of $2 to the power of p$ . $6 to the power of q$ . 15^r is-

2 to the power of p

6 to the power of q

15 to the power of r

- We need to find proper divisors.
Suppose

a to the power of n

is a number then factors of

a to the power of n

a to the power of 0 comma space a to the power of 1 comma space a squared comma a cubed comma........ space a to the power of n right parenthesis

and a is proper
i.e.

a to the power of n

has total division = (n + 1)
Now,

2 to the power of p

cross times

6 to the power of q

cross times 15 to the power of r

2 to the power of p cross times 2 to the power of q cross times 3 to the power of q cross times 3 to the power of r cross times 5 to the power of r

x to the power of m cross times space x to the power of n space equals x to the power of m space plus space n end exponent

2 to the power of p

cross times

6 to the power of q

cross times 15 to the power of r

2 to the power of p cross times 2 to the power of q cross times 3 to the power of q cross times 3 to the power of r cross times 5 to the power of r

2 to the power of left parenthesis p plus q right parenthesis end exponent cross times 3 to the power of left parenthesis q plus r right parenthesis end exponent cross times 5 to the power of r

Total factors = (p+q+1)(q+r+1)(r+1)
However, proper divisors exclude

1

and the number itself.
Hence, the answer is

(p + q + 1) (q + r + 1) (r + 1) - 2 .

The number of proper divisors of $2 to the power of p$ . $6 to the power of q$ . 15^r is-

2 to the power of p

6 to the power of q

15 to the power of r

- We need to find proper divisors.
Suppose

a to the power of n

is a number then factors of

a to the power of n

a to the power of 0 comma space a to the power of 1 comma space a squared comma a cubed comma........ space a to the power of n right parenthesis

and a is proper
i.e.

a to the power of n

has total division = (n + 1)
Now,

2 to the power of p

cross times

6 to the power of q

cross times 15 to the power of r

2 to the power of p cross times 2 to the power of q cross times 3 to the power of q cross times 3 to the power of r cross times 5 to the power of r

x to the power of m cross times space x to the power of n space equals x to the power of m space plus space n end exponent

2 to the power of p

cross times

6 to the power of q

cross times 15 to the power of r

2 to the power of p cross times 2 to the power of q cross times 3 to the power of q cross times 3 to the power of r cross times 5 to the power of r

2 to the power of left parenthesis p plus q right parenthesis end exponent cross times 3 to the power of left parenthesis q plus r right parenthesis end exponent cross times 5 to the power of r

Total factors = (p+q+1)(q+r+1)(r+1)
However, proper divisors exclude

1

and the number itself.
Hence, the answer is

(p + q + 1) (q + r + 1) (r + 1) - 2 .

If $x squared minus 11 x plus a text and end text x squared minus 14 x plus 2 a$ have a common factor then 'a' is equal to

If $x squared minus 11 x plus a text and end text x squared minus 14 x plus 2 a$ have a common factor then 'a' is equal to

A block C of mass is moving with velocity and collides elastically with block of mass and connected to another block of mass through spring constant .What is if is compression of spring when velocity of is same ?

Using conservation of linear momentum, we
have

Or
Using conservation of energy, we have

Where compression in the spring

Or

A block C of mass is moving with velocity and collides elastically with block of mass and connected to another block of mass through spring constant .What is if is compression of spring when velocity of is same ?

physics-General

Using conservation of linear momentum, we
have

Or
Using conservation of energy, we have

Where compression in the spring

Or

If $fraction numerator 1 over denominator left parenthesis 1 plus 2 x right parenthesis open parentheses 1 minus x squared close parentheses end fraction equals fraction numerator A over denominator 1 plus 2 x end fraction plus fraction numerator B over denominator 1 plus x end fraction plus fraction numerator C over denominator 1 minus x end fraction$ then assending order of A,B,C

If $fraction numerator 1 over denominator left parenthesis 1 plus 2 x right parenthesis open parentheses 1 minus x squared close parentheses end fraction equals fraction numerator A over denominator 1 plus 2 x end fraction plus fraction numerator B over denominator 1 plus x end fraction plus fraction numerator C over denominator 1 minus x end fraction$ then assending order of A,B,C

The number of different seven digit numbers that can be written using only the three digits 1, 2 and 3 with the condition that the digit 2 occurs twice in each number is-

Complete step by step solution:
We are given the number of different seven-digit numbers that can be written using only three digits 1,2 and 3. Therefore,
Total number of Digits

= 7

We are given that the digit two occurs exactly twice in each number.
Thus, the digit two occurs twice in the seven digit number.
Now, we will find the number of ways of arrangement of the digit two in the seven digit number by using combination.
Total number of ways that the digit two occurs exactly twice in each number

=^{7} C_{2}

Now, the remaining five digits can be written using two digits 1 and 3 in

2 to the power of 5

ways.
We will now find the total number of seven digit number by multiplying the number of ways of arrangement in both the cases. Therefore
Total number of seven digit number =

C presuperscript 7 subscript 2 space cross times space 2 to the power of 5

Now by using the formula

C presuperscript n subscript r space equals space begin inline style fraction numerator n factorial over denominator r factorial space left parenthesis n minus r right parenthesis factorial end fraction end style end subscript

, we get
Total number of seven digit number =

fraction numerator 7 factorial over denominator left parenthesis 7 minus 2 right parenthesis factorial space 2 factorial end fraction cross times space 2 to the power of 5

We know that the factorial can be written by the formula n! = n

cross times

(n-1)! , so we get
Total number of seven digit number =

fraction numerator 7 space cross times 6 cross times 5 factorial over denominator 5 factorial space 2 factorial end fraction cross times 2 to the power of 5

Total number of seven digit number =

fraction numerator 7 space cross times 6 over denominator 2 factorial end fraction cross times 2 to the power of 5

Simplifying the expression, we get
Total number of seven digit number

= 7 \times 6 \times 2^{4}

= 7 \times 6 \times 2^{4}

Total number of seven digit number

= 672

Therefore, the number of different seven-digit numbers that can be written using only three digits 1,2 and 3 is 672.

The number of different seven digit numbers that can be written using only the three digits 1, 2 and 3 with the condition that the digit 2 occurs twice in each number is-

Complete step by step solution:
We are given the number of different seven-digit numbers that can be written using only three digits 1,2 and 3. Therefore,
Total number of Digits

= 7

We are given that the digit two occurs exactly twice in each number.
Thus, the digit two occurs twice in the seven digit number.
Now, we will find the number of ways of arrangement of the digit two in the seven digit number by using combination.
Total number of ways that the digit two occurs exactly twice in each number

=^{7} C_{2}

Now, the remaining five digits can be written using two digits 1 and 3 in

2 to the power of 5

ways.
We will now find the total number of seven digit number by multiplying the number of ways of arrangement in both the cases. Therefore
Total number of seven digit number =

C presuperscript 7 subscript 2 space cross times space 2 to the power of 5

Now by using the formula

C presuperscript n subscript r space equals space begin inline style fraction numerator n factorial over denominator r factorial space left parenthesis n minus r right parenthesis factorial end fraction end style end subscript

, we get
Total number of seven digit number =

fraction numerator 7 factorial over denominator left parenthesis 7 minus 2 right parenthesis factorial space 2 factorial end fraction cross times space 2 to the power of 5

We know that the factorial can be written by the formula n! = n

cross times

(n-1)! , so we get
Total number of seven digit number =

fraction numerator 7 space cross times 6 cross times 5 factorial over denominator 5 factorial space 2 factorial end fraction cross times 2 to the power of 5

Total number of seven digit number =

fraction numerator 7 space cross times 6 over denominator 2 factorial end fraction cross times 2 to the power of 5

Simplifying the expression, we get
Total number of seven digit number

= 7 \times 6 \times 2^{4}

= 7 \times 6 \times 2^{4}

Total number of seven digit number

= 672

Therefore, the number of different seven-digit numbers that can be written using only three digits 1,2 and 3 is 672.

The centre and radius of the circle $r equals c o s space theta minus s i n space theta$ are respectively

The centre and radius of the circle $r equals c o s space theta minus s i n space theta$ are respectively

The centre of the circle $r squared minus 2 r left parenthesis 3 c o s space theta plus 4 s i n space theta right parenthesis minus 24$ is

The centre of the circle $r squared minus 2 r left parenthesis 3 c o s space theta plus 4 s i n space theta right parenthesis minus 24$ is