Biology
General
Easy

Question

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

  1. A = Eosinophil, B = Erythrocyte, C = Neutrophil and D = Basophil    
  2. A = Eosinophil, B = Lymphocyte, C = Neutrophil and D = Monocyte    
  3. A = Erythrocyte, B = Basophil, C = Neutrophil and D = Lymphocyte    
  4. A = Eosinophil, B = Monocyte, C = Neutrophil and D = Lymphocyte    

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

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The number of arrangements that can be formed by taking all the letters left curly bracket stack bold T comma bold U with _ below comma bold E comma bold S comma bold D comma bold A comma bold Y right curly bracket is:

The number of arrangements that can be formed by taking all the letters left curly bracket stack bold T comma bold U with _ below comma bold E comma bold S comma bold D comma bold A comma bold Y right curly bracket = P presuperscript 7 subscript 7 equals fraction numerator 7 factorial over denominator left parenthesis 7 minus 7 right parenthesis factorial end fraction equals fraction numerator 5040 over denominator 0 factorial end fraction equals 5040 over 1 equals 5040

The number of arrangements that can be formed by taking all the letters left curly bracket stack bold T comma bold U with _ below comma bold E comma bold S comma bold D comma bold A comma bold Y right curly bracket is:

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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?

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

No, of letters in the word FLOWER =6
No. of permutations that can be made using all the letters of the word FLOWER = P presuperscript 6 subscript 6 equals fraction numerator 6 factorial over denominator left parenthesis 6 minus 6 right parenthesis factorial end fraction equals fraction numerator 6 factorial over denominator 0 factorial end fraction equals 720 over 1 equals 720

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

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No, of letters in the word FLOWER =6
No. of permutations that can be made using all the letters of the word FLOWER = P presuperscript 6 subscript 6 equals fraction numerator 6 factorial over denominator left parenthesis 6 minus 6 right parenthesis factorial end fraction equals fraction numerator 6 factorial over denominator 0 factorial end fraction equals 720 over 1 equals 720
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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:

As we have to fill 8 places from 12 balls numbered 1 to 12.
So, the number of ways = P presuperscript 12 subscript 8

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:

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As we have to fill 8 places from 12 balls numbered 1 to 12.
So, the number of ways = P presuperscript 12 subscript 8
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If  left parenthesis n plus 1 right parenthesis P subscript 5 colon to the power of n P subscript 5 equals 3 colon 2then n is:

Given, left parenthesis n plus 1 right parenthesis P subscript 5 colon to the power of n P subscript 5 equals 3 colon 2
rightwards double arrow fraction numerator begin display style fraction numerator left parenthesis n plus 1 right parenthesis factorial over denominator left parenthesis n plus 1 minus 5 right parenthesis factorial end fraction end style over denominator begin display style fraction numerator n factorial over denominator left parenthesis n minus 5 right parenthesis factorial end fraction end style end fraction equals 3 over 2
rightwards double arrow fraction numerator left parenthesis n plus 1 right parenthesis factorial over denominator left parenthesis n minus 4 right parenthesis factorial end fraction cross times fraction numerator left parenthesis n minus 5 right parenthesis factorial over denominator n factorial end fraction equals 3 over 2
rightwards double arrow fraction numerator left parenthesis n plus 1 right parenthesis n factorial over denominator left parenthesis n minus 4 right parenthesis left parenthesis n minus 5 right parenthesis factorial end fraction cross times fraction numerator left parenthesis n minus 5 right parenthesis factorial over denominator n factorial end fraction equals 3 over 2
rightwards double arrow fraction numerator n plus 1 over denominator n minus 4 end fraction equals 3 over 2
rightwards double arrow 2 left parenthesis n plus 1 right parenthesis equals 3 left parenthesis n minus 4 right parenthesis
rightwards double arrow 2 n plus 2 equals 3 n minus 12
rightwards double arrow n equals 14

If  left parenthesis n plus 1 right parenthesis P subscript 5 colon to the power of n P subscript 5 equals 3 colon 2then n is:

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Given, left parenthesis n plus 1 right parenthesis P subscript 5 colon to the power of n P subscript 5 equals 3 colon 2
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rightwards double arrow fraction numerator left parenthesis n plus 1 right parenthesis factorial over denominator left parenthesis n minus 4 right parenthesis factorial end fraction cross times fraction numerator left parenthesis n minus 5 right parenthesis factorial over denominator n factorial end fraction equals 3 over 2
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rightwards double arrow fraction numerator n plus 1 over denominator n minus 4 end fraction equals 3 over 2
rightwards double arrow 2 left parenthesis n plus 1 right parenthesis equals 3 left parenthesis n minus 4 right parenthesis
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If blank to the power of n P subscript 5 equals to the power of n P subscript 6  then n is :

Given, blank to the power of n P subscript 5 equals to the power of n P subscript 6
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rightwards double arrow open parentheses n minus 6 close parentheses factorial equals open parentheses n minus 5 close parentheses factorial
rightwards double arrow open parentheses n minus 6 close parentheses factorial equals open parentheses n minus 5 close parentheses open parentheses n minus 6 close parentheses factorial
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If blank to the power of n P subscript 5 equals to the power of n P subscript 6  then n is :

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rightwards double arrow open parentheses n minus 6 close parentheses factorial equals open parentheses n minus 5 close parentheses factorial
rightwards double arrow open parentheses n minus 6 close parentheses factorial equals open parentheses n minus 5 close parentheses open parentheses n minus 6 close parentheses factorial
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number of ways of putting 4 letters in the right envelope= 1
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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 :

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Total number of ways of putting 4 letters in 4 envelopes = 4! =24
Now, number of ways of putting only 1 letter in the right envelope= blank to the power of 4 C subscript 1 cross times 2 =8
number of ways of putting only 2 letters in the right envelopes =blank to the power of 4 C subscript 2=6
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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:

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 = blank cubed C subscript 1=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:

Maths-General
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 = blank cubed C subscript 1=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
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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.

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

The number of numbers having 3 in the unit place= 4! =24
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=3 cross times 24 plus 4 cross times 24 plus 5 cross times 24 plus 6 cross times 24 plus 7 cross times 24
=24(3+4+5+6+7)
=24cross times25=600
Similarly the sum of the digits of all the numbers in each of the other places=600
The required sum =600 cross times 10000 plus 600 cross times 1000 plus 600 cross times 100 plus 600 cross times 10 plus 600 cross times 1
=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:

Maths-General
The number of numbers having 3 in the unit place= 4! =24
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=3 cross times 24 plus 4 cross times 24 plus 5 cross times 24 plus 6 cross times 24 plus 7 cross times 24
=24(3+4+5+6+7)
=24cross times25=600
Similarly the sum of the digits of all the numbers in each of the other places=600
The required sum =600 cross times 10000 plus 600 cross times 1000 plus 600 cross times 100 plus 600 cross times 10 plus 600 cross times 1
=600(10000+1000+100+10+1)
=600(11111)
=6666600
parallel

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