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Since there are 10 different books out of which 4 is to be selected.

(i) When there is no restriction

No. of ways in which 4 books be selected = ¹⁰C₄

= 210 ways

(ii) Two particular books are always selected

Since two particular books are always selected, so ways of selecting 2 books from 8 are = ⁸C₂ ways

= 28 ways

(iii) Two particular books are never selected

Since two particular books are never selected so, ways of selecting 4 books from 8 are = ⁸C₄ ways

= 70 ways

Given,

∑^n+rC_r r∈(0,m)

We know : 

ⁿC_r + ⁿC_r-1 = ⁿ⁺¹C_{r ....}(1)

\(\sum_{r=0}^{m} \) ^n+rC_r = ⁿC₀ + ⁿ⁺¹C₁ + ⁿ⁺²C₂ + ⁿ⁺³C₃ + . . . . . . + ^n+mC_n+1

\(\sum_{r=0}^{m} \)^n+rC_r = ⁿ⁺¹C₀ + ⁿ⁺¹C₁ + ⁿ⁺²C₂ + ⁿ⁺³C₃ + . . . . . . + ^n+mC_n+1 

⇒ (ⁿC₀ = ⁿ⁺¹C₀)

Using equation (1),

 \(\sum_{r=0}^{m} \)^n+rC_r = ⁿ⁺²C₁ + ⁿ⁺²C₂ + ⁿ⁺³C₃ + . . . . . . + ^n+mC_n+1

 \(\sum_{r=0}^{m} \)^n+rC_r = ⁿ⁺³C₂ + ⁿ⁺³C₃ + . . . . . . + ^n+mC_n+1

Proceeding in the same way :

 \(\sum_{r=0}^{m} \)^n+rC_r = ^n+mC_m-1 + ^n+mC_m = ^n+m+1C_n+1

 \(\sum_{r=0}^{m} \)^n+rC_r = ^n+m+1C_n+1

Given that,

We need to accommodate 21 guests to two round tables which can accommodate 15 and 6 persons each. 

We need to select 6 members first and arrange 6 and 15 members accordingly in the respective tables. 

Let us assume the no. of ways of choosing 6 members to be N₁ 

⇒ N₁ = No. of ways of choosing 6 members out of 21 guests. 

⇒ N₁ = ²¹C₆ 

Now,

We need to arrange the 6 members in a round table. 

By fixing a guest at a single seat, 

We arrange the remaining 5 members. 

The arrangement is similar to that of arranging n people in n places which are n! ways to arrange. 

So, 

The total no. of arrangements that can be made is 5!. 

Now,

We need to arrange the 15 members in a round table. By fixing a guest at a single seat, 

We arrange the remaining 5 members. 

The arrangement is similar to that of arranging n people in n places which are n! ways to arrange. 

So, 

The total no. of arrangements that can be made is 14!. 

Let us assume total no. of ways of arranging the guests in the table be N 

⇒ N = N₁ × 5! × 14! 

⇒ N = ²¹C₆ × 5! × 14! 

∴ The no. of ways of accommodating guests is ²¹C₆ × 5! × 14!.

As we know that if ⁿC_p = ⁿC_q, now one of the following conditions need to be satisfied:

(i) p = q

(ii) n = p + q

Therefore from the question ⁿC₄ = ⁿC₆, we can say that

4 ≠ 6

Therefore, the condition (ii) must be satisfied,

n = 4 + 6

n = 10

Then, we need to find ¹²C_n,

As we know that the value of n so, ¹²C_n = ¹²C₁₀

Lets use the formula,

ⁿC_r = n!/r!(n – r)!

Therefore now, value of n = 12 and r = 10

ⁿC_r = n!/r!(n – r)!

¹²C₁₀ = 12! / 10!(12 – 10)!

= 12! / (10! 2!)

= [12 × 11 × 10!] / (10! 2!)

= [12 × 11] / (2)

= 6 × 11

= 66

∴ The value of ¹²C₁₀ = 66.

As we know that if ⁿC_p = ⁿC_q, now one of the following conditions need to be satisfied:

(i) p = q

(ii) n = p + q

Therefore from the question ⁿC₁₂ = ⁿC₅, we can say that

12 ≠ 5

Therefore, the condition (ii) must be satisfied,

n = 12 + 5

n = 17

∴ The value of n is 17.

Option : (A)

n = 12 + 8 

(ⁿC_x = ⁿC_y ⇒ n = x + y or x = y) 

n = 20

Given: ⁿC_r+1 = ⁿC₈

Need to find: Value of ²²C_n

We know,

One of the property of combination is: If ⁿC_r = ⁿC_t, then,

(i) r = t OR

(ii) r + t = n We are going to use property

(i) ⁿC_r+1 = ⁿC₈ By the property (i),

⇒ r + 1 = 8

⇒ r = 7 Now we are going to use property

(ii) ⇒ n = 8 + 7 + 1 = 16

Therefore,

²²C_n = ²²C₁₆ = 74613.

Given,

¹⁵C_3r = ¹⁵C_r+3

We know that,

If ⁿC_P = ⁿC_q,

Then one of the following conditions need to be satisfied :

i. p = q 

ii. n = p + q 

Let us use condition (i),

⇒ 3r = r + 3 

⇒ 2r = 3

⇒ r = \(\frac{3}{2}\)

⇒ r + 3 = \(\frac{3}{2}\) + 3

⇒  r + 3 = \(\frac{9}{2}\)

We know that,

For a combination ⁿC_r,r≥0 and r should be an integer, which is not satisfies here, 

So, the condition (ii) must be satisfied, 

⇒ 15 = 3r + r + 3 

⇒ 4r = 12

⇒ r = \(\frac{12}{4}\) 

⇒ r = 4

∴ The value of r is 4.

Option : (A)

Three persons can take 5 seats in ⁵C₃ ways. 

Also,

3 persons can sit in 3! ways. 

∴ Required number of ways = ⁵C₃ x 3! 

= 10 x 6 

= 60

Total number of ways in which the letters can be put = 3! = 6 

Suppose, out of the three letters, one has been put in the correct envelope. 

This can be done in ³C₁ ways. (3 ways) 

Now, out of three, 

If two letters have been put in the current envelope, then the last one has been put in the correct envelope as well. 

This can be done in ³C₃ ways. (1 way) 

Number of ways = 3 + 1 = 4 

Number of ways in which no letter is put in correct envelope = 6 – 4 = 2

Given as

The total number of students in XI = 20

The total number of students in XII = 20

The total number of students to be selected in a team = 11 (with atleast 5 from class XI and 5 from class XII)

The number of ways = (No. of ways of selecting 6 students from class XI and 5 students from class XII) + (No. of ways of selecting 5 students from class XI and 6 students from class XII)

= (²⁰C₆ × ²⁰C₅) + (²⁰C₅ × ²⁰C₆)

= 2 (²⁰C₆ × ²⁰C₅) ways

There are 6 letters in the word ‘MONDAY’, and there is no letter repeating.

(i) 4 letters are used at first. 4 letters can sit in different ways.

So, here permutation is to be used.

So, the number of words that can be formed = ⁶P₄ = 360.

(ii) Now all the letters are used. Therefore, the number of words can be formed is = 6! = 720

(iii) Now the first letter is a vowel. There are 2 vowels in the word ‘MONDAY’, ‘O’ and ‘A’.

Let’s take ‘O’ as the first letter. Then we can place the 5 letters among the 5 places.

So, taking ‘O’ as the first letter, a number of words can be formed is = 5! = 120.

Similarly, taking ‘A’ as the first letter, a number of words can be formed = 5! = 120.

So, the total number of words can be formed taking first letter a vowel is = (120 + 120) = 240.

Given as

The total persons = 10

The number of persons to be selected = 5 from 10 persons (P₁, P₂, P₃ … P₁₀)

It is also told that P₁ should be present and P₄ and P₅ should not be present.

Now, we have to choose 4 persons from remaining 7 persons as P₁ is selected and P₄ and P₅ are already removed.

The number of ways = Selecting 4 persons from remaining 7 persons

= ⁷C₄

By using the formula,

ⁿC_r = n!/r!(n – r)!

⁷C₄ = 7! / 4!(7 – 4)!

= 7! / (4! 3!)

= [7 × 6 × 5 × 4!] / (4! 3!)

= [7 × 6 × 5] / (3 × 2 × 1)

= 7 × 5

= 35

Then we need to arrange the chosen 5 people. Since 1 person differs from other.

35 × 5! = 35 × (5 × 4 × 3 × 2 × 1)

= 4200

Hence, the total no. of possible arrangement can be done is 4200.

To get a straight line we just need to join two points. There are 12 numbers of points.

Therefore, there is ¹²C₂ = 66 number of straight lines. Among the 12 points, there are 3 points which are collinear.

That means joining those 3 lines give a single straight line.

That means the real number of straight lines present in the table is = (66 – ³C₂ + 1) = (66 – 3 + 1) = 64.

Given,

¹⁶C_r = ¹⁶C_{r + 2}

We know that,

If ⁿC_p = ⁿC_q,

Then one of the following conditions need to be satisfied :

i. p = q 

ii. n = p + q

From the problem,

¹⁶C_r = ¹⁶C_{r + 2}

We can say that,

⇒ r ≠ r + 2 

So, the condition(ii) must be satisfied, 

⇒ 16 = r + r + 2 

⇒ 2r = 14 

⇒ r = \(\frac{14}{2}\)

⇒ r = 7. 

∴ The value of r is 7.

Let us assume the negative consecutive integers are –1, – 2,......, – (2n) 

Let M be the product of the negative integers, 

⇒ M = ( – 1).( – 2)( – 3)......( – 2n + 1).( – 2n) 

⇒ M = ( – 1)2n(1.2.3.......(2n – 1).(2n)) 

⇒ M = 1.2.3......……(2n – 1).(2n) 

We know that, 

n! = n(n – 1)(n – 2)…………2.1 

⇒ M = (2n)!

⇒ \(\frac{M}{(2n)!}\) = \(\frac{(2n)!}{(2n)!}\) 

⇒ \(\frac{M}{(2n)!}\) = 1

∴ M is divisible by (2n)!.

An n-sided polygon has n vertices. 

By joining any two vertices of the polygon, we obtain either a side or a diagonal of the polygon. 

Number of line segments obtained by joining the vertices of an n-sided polygon if we take two vertices at a time, number of 

Ways of selection 2 out of n = ⁿC₂ 

Out of these lines, n lines are sides of the polygon. Number of diagonals of the polygon =

ⁿC₂ – n = \(\frac{n(n-1)}{2}\) - n

= \(\frac{n(n-3)}{2}\) 

n-sided polygon has n numbers of vertices. Diagonals are formed by joining the opposite vertices from one vertex, except the two adjacent vertices.

So, from one vertex (n - 3) diagonals can be drawn.

Similarly, for n numbers of vertices, n(n - 3) diagonals can be drawn. But, the diagonal joins 2 points at a time, here two vertices.

Therefore, the actual number of diagonals is \(=\frac{n(n-1)}{2}.\)

(i) a hexagon

As we know that a hexagon has 6 angular points. By joining those any two angular points we get a line which is either a side or a diagonal.

Therefore number of lines formed = ⁶C₂

On using the formula,

ⁿC_r = n!/r!(n – r)!

⁶C₂ = 6!/2!(6 - 2)!

= 6! / (2! 4!)

= [6 × 5 × 4!] / (2! 4!)

= [6 × 5] / (2 × 1)

= 3 × 5

= 15

As we know that number of sides of hexagon is 6

Therefore, number of diagonals = 15 – 6 = 9

Total no. of diagonals formed is 9.

(ii) a polygon of 16 sides

As we know that a polygon of 16 sides has 16 angular points. By joining those any two angular points we get a line which is either a side or a diagonal.

Therefore number of lines formed = ¹⁶C₂

On using the formula,

ⁿC_r = n!/r!(n – r)!

¹⁶C₂ = 16!/2!(16 - 2)!

= 16! / (2! 14!)

= [16 × 15 × 14!] / (2! 14!)

= [16 × 15] / (2 × 1)

= 8 × 15

= 120

As we know that number of sides of a polygon is 16

Therefore, number of diagonals = 120 – 16 = 104

Total no. of diagonals formed is 104.

Given as

The total number of questions = 5

The total number of questions to be answered = 4

The number of ways = we need to answer 2 questions out of the remaining 3 questions as 1 and 2 are compulsory.

= ³C₂

On using the formula,

ⁿC_r = n!/r!(n – r)!

³C₂ = 3!/2!(3 - 2)!

= 3! / (2! 1!)

= [3 × 2 × 1] / (2 × 1)

= 3

Hence, the no. of ways answering the questions is 3.

Option : (B)

r + r + 4 = 20 

(ⁿC_x = ⁿC_y ⇒ n = x + y or x = y ) 

2r + 4 = 20 

2r = 16 

r = 8 

Now,

^rC₃ = ⁸C₃

⁸C₃ = \(\frac{8!}{3!5!}\)

= \(\frac{8\times 7\times 6}{3\times 2\times 1}\)

= 56

Given: ^(n2–n)C₂ = ^(n2–n)C₄ = 120

Need to find: Value of n 

^(n2–n)C₂ = ^(n2–n)C₄ = 120

We know,

One of the property of combination is: If ⁿC_r = ⁿC_t, then,

(i) r = t OR

(ii) r + t = n

Applying property

(ii) we get, n² – n = 2 + 4 = 6

n² – n – 6 = 0

n² – 3n + 2n – 6 = 0

n(n – 3) + 2(n – 3) = 0

(n – 3) (n + 2) = 0

So, the value of n is either 3 or -2.

Option : (C)

A host lady can invite 8 out of 12 people in ¹²C₈ ways. 

Two out of these 12 people do not want to attend the party together. 

Number of ways = ¹²C₈ - ¹⁰C₆

Option : (B)

a² – a = 2 + 4 

(ⁿC_x = ⁿC_y ⇒ n = x + y or x = y) 

a² – a – 6 = 0 

a² – 3a – 2a – 6 = 0 

a(a – 3) + 2(a – 3) = 0 

(a + 2) (a – 3) = 0 

a = -2 or a = 3 

But, 

a = -2 is not possible since it’s negative. 

So, 

a = 3.

Option : (C)

5 out of 14 players are bowlers. 

In other words, 

9 players are not bowlers. 

A team of 11 is to be selected so as to include at least 4 bowlers. 

Number of ways = ⁵C₄ x  ⁹C₇ + ⁵C₅ x ⁹C₆ 

= 180 + 84 

= 264

Option : (B)

Number of straight lines formed by joining the 10 points if we take 2 points at a time = ¹⁰C₂

= \(\frac{10}{2}\times \frac{9}{1}\)

= 45

Number of straight lines formed by joining the 4 points if we take 2 points at a time = ⁴C₂

 = \(\frac{4}{2}\times \frac{3}{1}\)

= 6

But,

4 collinear points, when joined in pairs give only one line. 

∴ Required number of straight lines = 45 – 6 + 1 

= 40

Option : (B)

If set S has n elements, then C(n,k) is the number of ways of choosing k elements from S. 

Thus,

The number of subsets of S of all possible values is given by, 

C(n,0) + C(n,1) + C(n,2) + . . . . . . . . . . + C(n,n) = 2ⁿ

Comparing the given equation with the above equation we get,

2ⁿ = 256 

2ⁿ = 28 

n = 8 

∴ ²ⁿC₂ = ¹⁶C₂

= \(\frac{16!}{2!14!}\) 

= \(\frac{16\times 15}{2}\)

= 120

Option : (B)

4 out of 13 players are bowlers. 

In other words, 

9 players are not bowlers. 

A team of 11 is to be selected so as to include at least 2 bowlers. 

∴ Number of ways = ⁴C₂ x ⁹C₉ + ⁴C₃ x ⁹C₈ + ⁴C₄ x ⁹C₇ 

= 6 + 36 + 36 

= 78.

2 consonants out of 21 consonants can be chosen in ²¹C₂ ways.

3 vowels out of 5 vowels can be chosen in ⁵C₃ ways.

Length of the word is = (2 + 3) = 5 And also 5 letters can be written in 5! Ways.

Therefore, the number of words can be formed is = (²¹C₂ X ⁵C₃ X 5!) = 252000.

A parallelogram can be formed by choosing two parallel lines from the set of four parallel lines and two 

Parallel lines from the set of three parallel lines. 

Two parallel lines from the set of four parallel lines can be chosen in ⁴C₂ ways. 

Two parallel lines from the set of three parallel lines can be chosen in ³C₂ ways. 

Number of parallelograms that can be formed = ⁴C₂ \(\times\) ³C₂

= \(\frac{4!}{2!2!}\)\(\times \frac{3!}{2!1!}\)

= 6\(\times\)3 

= 18

It is clear that 3 things are already selected and we need to choose (r – 3) things from the remaining (n – 3) things.

Lets find the no. of ways of choosing (r – 3) things.

The number of ways = (No. of ways of choosing (r – 3) things from remaining (n – 3) things)

= ^{n – 3}C_{r – 3}

We need to find the no. of permutations than can be formed using 3 things which are together. Therefore, the total no. of words that can be formed is 3!

Let us assume the together things as a single thing this gives us total (r – 2) things which were present now. Therefore, the total no. of words that can be formed is (r – 2)!

The total number of words formed is:

^{n – 3}C_{r – 3} × 3! × (r – 2)!

Hence, the no. of permutations that can be formed by r things which are chosen from n things in which 3 things are always together is ^{n – 3}C_{r – 3} × 3! × (r – 2)!

⇒ ⁵C₁ + ⁵C₂ + ⁵C₃ + ⁵C₄ + ⁵C₅

⇒ ⁶C₂ + ⁶C₄ + 1 [As ⁵C₅ = 1]

⇒ 15 + 15 + 1

⇒ 31