Question

There are 3 letters and 3 addressed envelopes corresponding to them. The number of ways in which the letters be placed in the envelopes so that no letter is in the right envelope is:

- 5
- 3
- 1
- 2

Hint:

### First we will find the total number of ways of putting letters in the envelope, then we will find the number of ways of putting 1 letter, 2 letters and 3 letters in the right envelope respectively. At last, to find the total no, of ways when no letter will be in the right envelope we will subtract the number of ways of putting 1 letter, 2 letters and 3 letters in the right envelope respectively from total number of ways.

## The correct answer is: 2

### Total number ways in which 3 letters can be put in 3 different envelopes= 3! = 6

now, numbers of ways when only 1 letter is put in the right envelope = =3

number of ways when 2 letters are put in the right envelope = 0

as when we put 2 letters in the right envelope the third letter would be automatically in the right envelope

number of ways when 3 letters are put in the right envelope = 1

So, the no. of ways when no letter will be in the right envelope=6-(3+0+1) =6-4 =2

### Related Questions to study

### Match Column - I with Column - II and select the correct option from the codes give below.

### Match Column - I with Column - II and select the correct option from the codes give below.

The sum of all the numbers formed by taking all the digits from 3,4,5,6,7 is:

The number of numbers having 4 in the unit place= 4! =24

The number of numbers having 5 in the unit place= 4! =24

The number of numbers having 6 in the unit place= 4! =24

The number of numbers having 7 in the unit place= 4! =24

So the sum of the digits in the unit place of all the numbers=

=24(3+4+5+6+7)

=2425=600

Similarly the sum of the digits of all the numbers in each of the other places=600

The required sum =

=600(10000+1000+100+10+1)

=600(11111)

=6666600

The sum of all the numbers formed by taking all the digits from 3,4,5,6,7 is:

The number of numbers having 4 in the unit place= 4! =24

The number of numbers having 5 in the unit place= 4! =24

The number of numbers having 6 in the unit place= 4! =24

The number of numbers having 7 in the unit place= 4! =24

So the sum of the digits in the unit place of all the numbers=

=24(3+4+5+6+7)

=2425=600

Similarly the sum of the digits of all the numbers in each of the other places=600

The required sum =

=600(10000+1000+100+10+1)

=600(11111)

=6666600

The sum of all the numbers formed by taking all the digits from 2,3,4,5 is:

The number of numbers having 3 in the unit place= 3! =6

The number of numbers having 4 in the unit place= 3! =6

The number of numbers having 5 in the unit place= 3! =6

So the sum of the digits in the unit place of all the numbers=

=12+18+24+30

=84

Similarly the sum of the digits of all the numbers in each of the other places=84

The required sum =

=84(1000+100+10+1)

=84

=93324

The sum of all the numbers formed by taking all the digits from 2,3,4,5 is:

The number of numbers having 3 in the unit place= 3! =6

The number of numbers having 4 in the unit place= 3! =6

The number of numbers having 5 in the unit place= 3! =6

So the sum of the digits in the unit place of all the numbers=

=12+18+24+30

=84

Similarly the sum of the digits of all the numbers in each of the other places=84

The required sum =

=84(1000+100+10+1)

=84

=93324

### The number of constant mappings from to is

### The number of constant mappings from to is

### The number of many one functions from is

Element A can have an image in 3 ways, i.e., (1, 2, 3)

Similarly, B can have an image in 4 ways.

Thus, total number of ways =4×4×4=64

### The number of many one functions from is

Element A can have an image in 3 ways, i.e., (1, 2, 3)

Similarly, B can have an image in 4 ways.

Thus, total number of ways =4×4×4=64

### The number of into functions that can be defined from is

### The number of into functions that can be defined from is

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A body moving with velocity v has momentum and kinetic energy numerically equal. What is the value of v?

The graph between the resistive force F acting on a body and the distance covered by the body is shown in figure. The mass of the body is 25KG and initial velocity is 2 m/s. When the distance covered by the body is 4km, its kinetic energy would be

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The number of onto functions that can be defined from A={a,b,c,d,e} to {1,2} is

Given, $A={a,b,c,d, e}$ and $B={1,2}$

Element $a$ can has an image in 2 ways, i.e., $(1$ or 2).

Similarly, a, $b,c, d$ and e can have an image in 5 ways.

Thus, total number of ways $=2$$×2×2×2×2=32$

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Given, $A={a,b,c,d, e}$ and $B={1,2}$

Element $a$ can has an image in 2 ways, i.e., $(1$ or 2).

Similarly, a, $b,c, d$ and e can have an image in 5 ways.

Thus, total number of ways $=2$$×2×2×2×2=32$

The momentum of a body increases by 20%. The percentage increase in its kinetic energy is

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