Question

# How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even position ?

- 16
- 36
- 60
- 180

Hint:

### Here we need to find the total number of nine digit numbers that can be formed using the given digits. We will count the number of even places present for the odd digits and then we will find the number of odd places present for the even digits. Then we will find the number of ways to arrange the odd digits and then we will find the number of ways to arrange the even digits and to get the final answer, we will multiply both of them.

## The correct answer is: 60

### Detailed Solution

Here we need to find the total number of nine digit numbers that can be formed using the given digits i.e. 2, 2, 3, 3, 5, 5, 8, 8, 8.

The digits which are even are 2, 2, 8, 8 and 8.

The digits which are odd are 3, 3, 5 and 5.

We have to arrange the odd digits in even places.

On finding the value of the factorials, we get

On further simplification, we get

Now, we have to arrange the even digits in odd places.

On finding the value of the factorials, we get

On further simplification, we get

Here we have obtained the total number of 9 digit numbers using the given digits. While finding the number of ways to arrange the odd digits in 5 even places, we have divided the 4! by 2! because the digit 3 were occurring two times and the digit 5 were occurring 2 times. Here we can make a mistake by conserving the number of even digits 4 and the number of odd digits 5, which will result in the wrong answer.

### Related Questions to study

### 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 :

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 =

Thus, total number of ways in which he can make the predictions so that exactly 10 predictions are correct, is equal to

### 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 :

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 =

Thus, total number of ways in which he can make the predictions so that exactly 10 predictions are correct, is equal to

### The foot of the perpendicular from the point on the line is

### The foot of the perpendicular from the point on the line is

### The point of intersection of the lines is

### The point of intersection of the lines is

### The line passing through and perpendicular to is

### The line passing through and perpendicular to is

### The equation of the line passing through is

### The equation of the line passing through is

### The length of the perpendicular from (-1, π/6) to the line is

### The length of the perpendicular from (-1, π/6) to the line is

### 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) :

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.

Now, we get a sum of digits in units place for all the numbers as 14×6=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.

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

(84×1000)+(84×100)+(84×10)+(84×1)

Sum = 84000+8400+840+84

Sum = 93324

We have found the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time as 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) :

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.

Now, we get a sum of digits in units place for all the numbers as 14×6=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.

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

(84×1000)+(84×100)+(84×10)+(84×1)

Sum = 84000+8400+840+84

Sum = 93324

We have found the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time as 93324.

### Total number of divisors of 480, that are of the form 4n + 2, n 0, is equal to :

To solve this question, we should know that the total number of divisors of any number x of the form are prime numbers and is given by (m + 1) (n + 1) (p + 1)….. we know that 480 can be expressed as

So, according to the formula, the total number of divisors of 480 are (5 + 1) (1 + 1) (1 + 1) = 6×2×2=24

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. So, the total number of odd divisors that are possible are (1 + 1) (1 + 1) = 2×2=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,

So, the number of divisors that are multiples of 4 are (3 + 1) (1 + 1) (1 + 1) = 4×2×2 = 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≥0

.

### Total number of divisors of 480, that are of the form 4n + 2, n 0, is equal to :

To solve this question, we should know that the total number of divisors of any number x of the form are prime numbers and is given by (m + 1) (n + 1) (p + 1)….. we know that 480 can be expressed as

So, according to the formula, the total number of divisors of 480 are (5 + 1) (1 + 1) (1 + 1) = 6×2×2=24

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. So, the total number of odd divisors that are possible are (1 + 1) (1 + 1) = 2×2=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,

So, the number of divisors that are multiples of 4 are (3 + 1) (1 + 1) (1 + 1) = 4×2×2 = 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≥0

.

### If ^{9}P_{5} + 5 ^{9}P_{4} = ^{10}P_{r }, then r =

Using Formula :

Dividing both sides by 9!

### If ^{9}P_{5} + 5 ^{9}P_{4} = ^{10}P_{r }, then r =

Using Formula :

Dividing both sides by 9!

### The number of proper divisors of . . 15^{r} is-

Suppose is a number then factors of = ( and a is proper

i.e. has total division = (n + 1)

Now, =

We know that

Thus, = =

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 . . 15^{r} is-

Suppose is a number then factors of = ( and a is proper

i.e. has total division = (n + 1)

Now, =

We know that

Thus, = =

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 have a common factor then 'a' is equal to

### If 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 ?

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 ?

have

Or

Using conservation of energy, we have

Where compression in the spring

Or

### If then ascending order of A, B, C.

### If then ascending 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-

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 =

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

Now by using the formula

, we get

Total number of seven digit number =

We know that the factorial can be written by the formula n! = n(n-1)! , so we get

Total number of seven digit number =

Total number of seven digit number =

Simplifying the expression, we get

Total number of seven digit number =

Multiplying the terms, we get

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-

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 =

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

Now by using the formula

, we get

Total number of seven digit number =

We know that the factorial can be written by the formula n! = n(n-1)! , so we get

Total number of seven digit number =

Total number of seven digit number =

Simplifying the expression, we get

Total number of seven digit number =

Multiplying the terms, we get

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.