Question

# Study the diagram given below and identify the cells labelled as A, B, C and D, and choose the correct option

- A = Eosinophil, B = Erythrocyte, C = Neutrophil and D = Basophil
- A = Eosinophil, B = Lymphocyte, C = Neutrophil and D = Monocyte
- A = Erythrocyte, B = Basophil, C = Neutrophil and D = Lymphocyte
- A = Eosinophil, B = Monocyte, C = Neutrophil and D = Lymphocyte

## The correct answer is: A = Eosinophil, B = Monocyte, C = Neutrophil and D = Lymphocyte

### Related Questions to study

### The number of arrangements that can be formed by taking all the letters is:

### The number of arrangements that can be formed by taking all the letters is:

### The accompanying diagram shows the three stages in the cardiac cycle.

Which of the following is the correct sequence?

### The accompanying diagram shows the three stages in the cardiac cycle.

Which of the following is the correct sequence?

No. of permutations that can be made using all the letters of the word FLOWER is :

No. of permutations that can be made using all the letters of the word FLOWER =

No. of permutations that can be made using all the letters of the word FLOWER is :

No. of permutations that can be made using all the letters of the word FLOWER =

There are 12 balls numbered from 1 to 12 The number of ways in which they can be used to fill 8 places in a row is:

So, the number of ways =

There are 12 balls numbered from 1 to 12 The number of ways in which they can be used to fill 8 places in a row is:

So, the number of ways =

### If then n is:

### If then n is:

### If then n is :

### If then n is :

### If then r is :

### If then r is :

### In the figure given below, which blood vessel represents vena cava?

### In the figure given below, which blood vessel represents vena cava?

### Match the types of WBC listed under Column - I with the shape of nucleus given under Column - II and select the correct option from codes given below.

### Match the types of WBC listed under Column - I with the shape of nucleus given under Column - II and select the correct option from codes given below.

The number of ways in which four letters can put in four addressed envelopes so that no letter goes into the envelope meant for it is :

Now, number of ways of putting only 1 letter in the right envelope= =8

number of ways of putting only 2 letters in the right envelopes ==6

number of ways of putting 3 letters in the right envelope= 0, as when we will put three letters in the right envelope the last will automatically go in the right place.

number of ways of putting 4 letters in the right envelope= 1

So, the number of ways in which four letters can put in four addressed envelopes so that no letter goes into the envelope meant for it =24-(8+6+0+1)=24-15=9

The number of ways in which four letters can put in four addressed envelopes so that no letter goes into the envelope meant for it is :

Now, number of ways of putting only 1 letter in the right envelope= =8

number of ways of putting only 2 letters in the right envelopes ==6

number of ways of putting 3 letters in the right envelope= 0, as when we will put three letters in the right envelope the last will automatically go in the right place.

number of ways of putting 4 letters in the right envelope= 1

So, the number of ways in which four letters can put in four addressed envelopes so that no letter goes into the envelope meant for it =24-(8+6+0+1)=24-15=9

### In the given figure of the heart which of the labelled part (1, 2, 3, 4, 5) carries oxygenated blood?

### In the given figure of the heart which of the labelled part (1, 2, 3, 4, 5) carries oxygenated blood?

### The given figure shows an angiogram of the coronary blood vessel. Which one of the following statements correctly describes, what is being done?

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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:

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

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:

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

### 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