Biology
General
Easy

Question

Which one is absent in free state during origin of life ?

  1. O subscript 2 end subscript    
  2. H subscript 2 end subscript    
  3. N subscript 2 end subscript    
  4. N H subscript 3 end subscript    

The correct answer is: O subscript 2 end subscript

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Two different packs of 52 cards are shuffled together. The number of ways in which a man can be dealt 26 cards so that he does not get two cards of the same suit and same denomination is-

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In how many ways can 6 prizes be distributed equally among 3 persons?

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The number of ways in which mn students can be distributed equally among m sections is-

 Detailed Solution
In this question, we have been asked to find the number of ways in which we can distribute mn students equally among m section. Now, we have been given that there are mn students in total and they have to be distributed equally among m sections. So, we can say in each section, there will be fraction numerator m n over denominator m end fraction space equals space n space s t u d e n t s.

Now, we take n students from mn students for each section by using the formula of combination, that is,
C presuperscript n subscript r space equals space fraction numerator n factorial over denominator r factorial left parenthesis n minus r right parenthesis factorial end fraction
For the first section, we can choose n students out of mn. So, we get the number of ways of choosing n students for the first section as C presuperscript m n end presuperscript subscript n space.
For the second section, we will again choose n students but out of (mn – n) because n students have already been chosen for the first section. So, we get the number of ways of choosing n students for the second section as C presuperscript m n minus n end presuperscript subscript n space
Similarly, for the third section, we have to choose n students out of (mn – 2n). So, we get the number of ways of choosing n students for the third section as C presuperscript m n minus 2 n end presuperscript subscript n space
And we will continue it in the same manner up to all mn students will not be divided into m section.
So, for (m – 1)th section, we will choose n students from (mn – ( m – 2)n) student. So, we get the number of ways of choosing n students for (n – 1)th section, we get, C presuperscript m n minus left parenthesis m minus 2 right parenthesis n end presuperscript subscript n space


And for the mth section, we get the number of ways for choosing students as, C presuperscript m n minus left parenthesis m minus 1 right parenthesis n end presuperscript subscript n space
Hence, we can write the total number of ways of distributing mn students in m section as 
C presuperscript m n end presuperscript subscript n space cross times space C presuperscript m n minus n end presuperscript subscript n space end subscript cross times space C presuperscript m n minus 2 n end presuperscript subscript n space end subscript cross times space......... space cross times space C presuperscript m n minus left parenthesis m minus 2 right parenthesis n end presuperscript subscript n space end subscript space cross times space C presuperscript m n minus left parenthesis m minus 1 right parenthesis n end presuperscript subscript n space end subscript
Now, we will use the formula of C presuperscript n subscript r space equals space fraction numerator n factorial over denominator r factorial left parenthesis n minus r right parenthesis factorial end fractionto expand it. So, we get,
space fraction numerator left parenthesis m n right parenthesis factorial over denominator n factorial left parenthesis m n minus n right parenthesis factorial end fraction space cross times space fraction numerator left parenthesis m n minus n right parenthesis factorial over denominator n factorial left parenthesis m n minus 2 n right parenthesis factorial end fraction space cross times space fraction numerator left parenthesis m n minus 2 n right parenthesis factorial over denominator n factorial left parenthesis m n minus 3 n right parenthesis factorial end fraction space cross times space..... space cross times space fraction numerator left parenthesis m n minus left parenthesis m minus 2 right parenthesis n right parenthesis factorial over denominator n factorial left parenthesis m n minus left parenthesis m minus 1 right parenthesis n right parenthesis factorial end fraction space cross times space fraction numerator left parenthesis m n minus left parenthesis m minus 1 right parenthesis n right parenthesis factorial over denominator n factorial left parenthesis m n minus m n right parenthesis factorial end fraction


And we can further write it as,
fraction numerator m n factorial over denominator n factorial end fraction space cross times fraction numerator 1 over denominator n factorial end fraction cross times space fraction numerator 1 over denominator n factorial end fraction space cross times.... cross times space 1 space equals space fraction numerator left parenthesis m n right parenthesis factorial over denominator left parenthesis n factorial right parenthesis to the power of m end fraction
Thus, we can say that the total number of ways of distributing mn students in m section are  fraction numerator left parenthesis m n right parenthesis factorial over denominator left parenthesis n factorial right parenthesis to the power of m end fraction

The number of ways in which mn students can be distributed equally among m sections is-

Maths-General
 Detailed Solution
In this question, we have been asked to find the number of ways in which we can distribute mn students equally among m section. Now, we have been given that there are mn students in total and they have to be distributed equally among m sections. So, we can say in each section, there will be fraction numerator m n over denominator m end fraction space equals space n space s t u d e n t s.

Now, we take n students from mn students for each section by using the formula of combination, that is,
C presuperscript n subscript r space equals space fraction numerator n factorial over denominator r factorial left parenthesis n minus r right parenthesis factorial end fraction
For the first section, we can choose n students out of mn. So, we get the number of ways of choosing n students for the first section as C presuperscript m n end presuperscript subscript n space.
For the second section, we will again choose n students but out of (mn – n) because n students have already been chosen for the first section. So, we get the number of ways of choosing n students for the second section as C presuperscript m n minus n end presuperscript subscript n space
Similarly, for the third section, we have to choose n students out of (mn – 2n). So, we get the number of ways of choosing n students for the third section as C presuperscript m n minus 2 n end presuperscript subscript n space
And we will continue it in the same manner up to all mn students will not be divided into m section.
So, for (m – 1)th section, we will choose n students from (mn – ( m – 2)n) student. So, we get the number of ways of choosing n students for (n – 1)th section, we get, C presuperscript m n minus left parenthesis m minus 2 right parenthesis n end presuperscript subscript n space


And for the mth section, we get the number of ways for choosing students as, C presuperscript m n minus left parenthesis m minus 1 right parenthesis n end presuperscript subscript n space
Hence, we can write the total number of ways of distributing mn students in m section as 
C presuperscript m n end presuperscript subscript n space cross times space C presuperscript m n minus n end presuperscript subscript n space end subscript cross times space C presuperscript m n minus 2 n end presuperscript subscript n space end subscript cross times space......... space cross times space C presuperscript m n minus left parenthesis m minus 2 right parenthesis n end presuperscript subscript n space end subscript space cross times space C presuperscript m n minus left parenthesis m minus 1 right parenthesis n end presuperscript subscript n space end subscript
Now, we will use the formula of C presuperscript n subscript r space equals space fraction numerator n factorial over denominator r factorial left parenthesis n minus r right parenthesis factorial end fractionto expand it. So, we get,
space fraction numerator left parenthesis m n right parenthesis factorial over denominator n factorial left parenthesis m n minus n right parenthesis factorial end fraction space cross times space fraction numerator left parenthesis m n minus n right parenthesis factorial over denominator n factorial left parenthesis m n minus 2 n right parenthesis factorial end fraction space cross times space fraction numerator left parenthesis m n minus 2 n right parenthesis factorial over denominator n factorial left parenthesis m n minus 3 n right parenthesis factorial end fraction space cross times space..... space cross times space fraction numerator left parenthesis m n minus left parenthesis m minus 2 right parenthesis n right parenthesis factorial over denominator n factorial left parenthesis m n minus left parenthesis m minus 1 right parenthesis n right parenthesis factorial end fraction space cross times space fraction numerator left parenthesis m n minus left parenthesis m minus 1 right parenthesis n right parenthesis factorial over denominator n factorial left parenthesis m n minus m n right parenthesis factorial end fraction


And we can further write it as,
fraction numerator m n factorial over denominator n factorial end fraction space cross times fraction numerator 1 over denominator n factorial end fraction cross times space fraction numerator 1 over denominator n factorial end fraction space cross times.... cross times space 1 space equals space fraction numerator left parenthesis m n right parenthesis factorial over denominator left parenthesis n factorial right parenthesis to the power of m end fraction
Thus, we can say that the total number of ways of distributing mn students in m section are  fraction numerator left parenthesis m n right parenthesis factorial over denominator left parenthesis n factorial right parenthesis to the power of m end fraction
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The number of ways to make 5 heaps of 3books each from 15 different books is-

Detailed Solution :
5 heaps of 3 books each are to be made from 15 different books. We are to find in how many ways this can be done.
We know that the formula of dividing different things into groups of sizes a subscript 1 comma space a subscript 2 comma a subscript 3 space end subscript........ a subscript n space where
a subscript 1 plus space a space plus space a subscript 3 space space end subscript plus space........ plus a subscript n space = fraction numerator m factorial over denominator left parenthesis a subscript 1 factorial right parenthesis left parenthesis a subscript 2 factorial right parenthesis left parenthesis a subscript 3 factorial right parenthesis......... left parenthesis a subscript n factorial right parenthesis end fraction
But this is applicable on if the groups are of unequal sizes.
In the given problem all the groups are of size 3, and there are 5 groups.
Hence,
a subscript 1 equals space a subscript 2 equals space a subscript 3 space end subscript equals a subscript 4 space end subscript space equals space a subscript 5 equals space 3


m  = 15

Again, the equal groups will be arranged amongst themselves, which is possible in 5ways. Thus, we can say that the required number of ways in which the 15 different books can be divided into 5 groups of size 3 is
fraction numerator 15 factorial over denominator left parenthesis 3 factorial right parenthesis left parenthesis 3 factorial right parenthesis left parenthesis 3 factorial right parenthesis left parenthesis 3 factorial right parenthesis left parenthesis 3 factorial right parenthesis space. left parenthesis space 5 factorial right parenthesis end fraction space equals space fraction numerator 15 factorial over denominator left parenthesis 3 factorial right parenthesis to the power of 5 space cross times space 5 factorial end fraction

The number of ways to make 5 heaps of 3books each from 15 different books is-

Maths-General
Detailed Solution :
5 heaps of 3 books each are to be made from 15 different books. We are to find in how many ways this can be done.
We know that the formula of dividing different things into groups of sizes a subscript 1 comma space a subscript 2 comma a subscript 3 space end subscript........ a subscript n space where
a subscript 1 plus space a space plus space a subscript 3 space space end subscript plus space........ plus a subscript n space = fraction numerator m factorial over denominator left parenthesis a subscript 1 factorial right parenthesis left parenthesis a subscript 2 factorial right parenthesis left parenthesis a subscript 3 factorial right parenthesis......... left parenthesis a subscript n factorial right parenthesis end fraction
But this is applicable on if the groups are of unequal sizes.
In the given problem all the groups are of size 3, and there are 5 groups.
Hence,
a subscript 1 equals space a subscript 2 equals space a subscript 3 space end subscript equals a subscript 4 space end subscript space equals space a subscript 5 equals space 3


m  = 15

Again, the equal groups will be arranged amongst themselves, which is possible in 5ways. Thus, we can say that the required number of ways in which the 15 different books can be divided into 5 groups of size 3 is
fraction numerator 15 factorial over denominator left parenthesis 3 factorial right parenthesis left parenthesis 3 factorial right parenthesis left parenthesis 3 factorial right parenthesis left parenthesis 3 factorial right parenthesis left parenthesis 3 factorial right parenthesis space. left parenthesis space 5 factorial right parenthesis end fraction space equals space fraction numerator 15 factorial over denominator left parenthesis 3 factorial right parenthesis to the power of 5 space cross times space 5 factorial end fraction
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The line among the following that touches the y to the power of 2 end exponent equals 4 a x is

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The number of necklaces which can be formed by selecting 4 beads out of 6 beads of different coloured glasses and 4 beads out of 5 beads of different metal, is-

The number of necklaces which can be formed by selecting 4 beads out of 6 beads of different coloured glasses and 4 beads out of 5 beads of different metal, is-

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The number of ways in which 20 persons can sit on 8 chairs round a circular table is-

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How many numbers consisting of 5 digits can be formed in which the digits 3,4 and 7 are used only once and the digit 5 is used twice-

How many numbers consisting of 5 digits can be formed in which the digits 3,4 and 7 are used only once and the digit 5 is used twice-

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The number of ways of distributing n prizes among n boys when any of the student does not get all the prizes is-

The number of ways of distributing n prizes among n boys when any of the student does not get all the prizes is-

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If the line x plus y plus k equals 0 is a tangent to the parabola x to the power of 2 end exponent equals 4 y then k=

If the line x plus y plus k equals 0 is a tangent to the parabola x to the power of 2 end exponent equals 4 y then k=

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The number of ways in which n prizes can be distributed among n students when each student is eligible to get any number of prizes is-

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In how many ways can six different rings be wear in four fingers?

Detailed Solution
Now, we have been given that there are 6 rings of different types and we have to find the ways in which they can be worn in 4 fingers.
Now, we know that the number of options each ring has is 4, that is each ring has 4 fingers as their possible way as it can be worn in any one of 4 fingers.
Now, similarly the other rings will have four options as it has not been mentioned in the options that there has to be at least a ring in a finger. So, each ring has four options i.e. four fingers.

Now, we know that by the fundamental principle of counting there can be = 4 x 4× 4× 4× 4× ways of wearing 6 rings.
So, we have 4 to the power of 6 = 4096 ways to wear 6 different types of rings.

In how many ways can six different rings be wear in four fingers?

Maths-General
Detailed Solution
Now, we have been given that there are 6 rings of different types and we have to find the ways in which they can be worn in 4 fingers.
Now, we know that the number of options each ring has is 4, that is each ring has 4 fingers as their possible way as it can be worn in any one of 4 fingers.
Now, similarly the other rings will have four options as it has not been mentioned in the options that there has to be at least a ring in a finger. So, each ring has four options i.e. four fingers.

Now, we know that by the fundamental principle of counting there can be = 4 x 4× 4× 4× 4× ways of wearing 6 rings.
So, we have 4 to the power of 6 = 4096 ways to wear 6 different types of rings.
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How many signals can be given by means of 10 different flags when at a time 4 flags are used, one above the other?

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