Biology

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

#### The correct answer is:

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### Related Questions to study

Maths-

#### 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-

#### 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-

Maths-General

Maths-

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

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

Maths-General

Maths-

#### 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

Now, we take n students from mn students for each section by using the formula of combination, that is,

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 .

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

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

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,

And for the mth section, we get the number of ways for choosing students as,

Hence, we can write the total number of ways of distributing mn students in m section as

Now, we will use the formula of r to expand it. So, we get,

And we can further write it as,

Thus, we can say that the total number of ways of distributing mn students in m section are

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

Now, we take n students from mn students for each section by using the formula of combination, that is,

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 .

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

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

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,

And for the mth section, we get the number of ways for choosing students as,

Hence, we can write the total number of ways of distributing mn students in m section as

Now, we will use the formula of r to expand it. So, we get,

And we can further write it as,

Thus, we can say that the total number of ways of distributing mn students in m section are

#### 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

Now, we take n students from mn students for each section by using the formula of combination, that is,

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 .

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

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

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,

And for the mth section, we get the number of ways for choosing students as,

Hence, we can write the total number of ways of distributing mn students in m section as

Now, we will use the formula of r to expand it. So, we get,

And we can further write it as,

Thus, we can say that the total number of ways of distributing mn students in m section are

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

Now, we take n students from mn students for each section by using the formula of combination, that is,

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 .

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

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

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,

And for the mth section, we get the number of ways for choosing students as,

Hence, we can write the total number of ways of distributing mn students in m section as

Now, we will use the formula of r to expand it. So, we get,

And we can further write it as,

Thus, we can say that the total number of ways of distributing mn students in m section are

Maths-

#### The number of ways in which 20 volunteers can be divided into groups of 4, 7 and 9 persons is-

#### The number of ways in which 20 volunteers can be divided into groups of 4, 7 and 9 persons is-

Maths-General

Maths-

#### 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

=

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,

m = 15

Again, the equal groups will be arranged amongst themselves, which is possible in 5! ways. 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

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

*m**different things into groups of sizes*where=

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,

m = 15

Again, the equal groups will be arranged amongst themselves, which is possible in 5! ways. 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

#### 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

=

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,

m = 15

Again, the equal groups will be arranged amongst themselves, which is possible in 5! ways. 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

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

*m**different things into groups of sizes*where=

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,

m = 15

Again, the equal groups will be arranged amongst themselves, which is possible in 5! ways. 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

Maths-

#### The line among the following that touches the is

#### The line among the following that touches the is

Maths-General

Maths-

#### 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-

Maths-General

Maths-

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

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

Maths-General

Maths-

#### The number of numbers can be formed by taking any 2 digits from digits 6,7,8,9 and 3 digits from 1, 2, 3, 4, 5 is -

#### The number of numbers can be formed by taking any 2 digits from digits 6,7,8,9 and 3 digits from 1, 2, 3, 4, 5 is -

Maths-General

Maths-

#### 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-

Maths-General

maths-

#### 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-

maths-General

Maths-

#### If the line is a tangent to the parabola then k=

#### If the line is a tangent to the parabola then k=

Maths-General

Maths-

#### 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-

#### 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-

Maths-General

Maths-

#### 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× 4 ways of wearing 6 rings.

So, we have = 4096 ways to wear 6 different types of rings.

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× 4 ways of wearing 6 rings.

So, we have = 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× 4 ways of wearing 6 rings.

So, we have = 4096 ways to wear 6 different types of rings.

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× 4 ways of wearing 6 rings.

So, we have = 4096 ways to wear 6 different types of rings.

Maths-

#### How many signals can be given by means of 10 different flags when at a time 4 flags are used, one above the other?

#### How many signals can be given by means of 10 different flags when at a time 4 flags are used, one above the other?

Maths-General