Maths-

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Easy

### Question

#### If a,b,c,d are consective binomial coefficients of then are is

- A.P
- G.P
- H.P
- A.G.P

#### The correct answer is: H.P

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### Related Questions to study

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No of terms whose value depend on 'x' is is

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If represent the terms is then is

If represent the terms is then is

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#### Let ' O be the origin and A be a point on the curve then locus of the midpoint of OA is

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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

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

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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

=

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-

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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

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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-

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#### 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 -

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