Combinatorics MCQ Quiz - Objective Question with Answer for Combinatorics - Download Free PDF

Calculation:

We have,

1/1⁴ + 1/2⁴ + 1/3⁴ + 1/4⁴ + ⋯ = π⁴ / 90         (1)

This can be written as:

[1/1⁴ + 1/3⁴ + 1/5⁴ + ⋯ ] + 1/2⁴ + 1/4⁴ + ⋯ = π⁴ / 90

[1/1⁴ + 1/3⁴ + 1/5⁴ + ⋯ ] + (1/2⁴) × [1/1⁴ + 1/2⁴ + 1/3⁴ + ⋯ ] = π⁴ / 90

Using (1):

[1/1⁴ + 1/3⁴ + 1/5⁴ + ⋯ ] + (1/2⁴) × (π⁴ / 90) = π⁴ / 90

⇒ 1/1⁴ + 1/3⁴ + 1/5⁴ + ⋯ = (π⁴ / 90) × [1 − 1/2⁴] = (π⁴ / 90) × (15/16) = π⁴ / 96

Given:

First term (a) = 27

Common ratio (r) = 2/3

Find the 4th term of the GP.

Formula used:

n-th term of GP = a × r^(n-1)

Calculation:

4th term = 27 × (2/3)^(4-1)

⇒ 4th term = 27 × (2/3)³

⇒ 4th term = 27 × (8/27)

⇒ 4th term = 8

∴ The correct answer is option (1).

The correct answer is : option 2.

Key Points

Explanation of the Recurrence Relation Solution

Given the recurrence relation: T(n) = T(2n/3) + 1

To solve this, we will use the Master Theorem for divide-and-conquer recurrences of the form:

T(n) = aT(n/b) + f(n)

In our case, a = 1, b = 3/2, and f(n) = 1.

Using the Master Theorem, we compare f(n) with n^log_b(a):

• a = 1
• b = 3/2
• log_b(a) = log(1) / log(3/2) = 0
• f(n) = 1

Since f(n) = 1 is Θ(1), it matches the case where f(n) = Θ(n^c) with c = 0. This corresponds to case 2 of the Master Theorem, where f(n) is polynomially smaller than n^log_b(a).

Therefore, the solution to the recurrence is:

T(n) = Θ(log n)

Thus, the correct answer is option 2.

\(x=\frac{1}{5+\frac{1}{6+x}}\)

\(\Rightarrow x=\frac{6+x}{31+5 x}\)

\(\Rightarrow 5 x^{2}+31 x=x+6\)

\(\Rightarrow 5 x^{2}+30 x-6=0\)

\(\Rightarrow x=\frac{-30 \pm \sqrt{900+120}}{10}=-3 \pm \sqrt{10.2}\)

The continued fraction has to be positive.

Therefore, We reject the negative value.

Concept:

Power series:

A series of the form \(\sum_{n = 0}^{\infty}a_n(z\ -\ z_0)^n\) is called power series about a point z = z₀.

If z₀ = 0, then power series about z₀ = 0 become \(\sum_{n = 0}^{\infty}a_nz^n\)

The radius of convergence: 

Let \(\sum_{n = 0}^{\infty}a_n(z\ -\ z_0)^n\)  be a power series and let R be the radius of circle in which power series is convergence and 

\(\frac{1}{R}\ =\ \mathop {\lim }\limits_{n \to {\rm{\infty}}}|a_n|^{\frac{1}[n}\ =\ \mathop {\lim }\limits_{n \to {\rm{\infty}}}|\frac{a_{n+1}}{a_n}|\)

Calculation:

Given power series is

\( \sum_{n\ =\ 1}^{\infty}[\frac{7^n}{n(5x\ -\ 1)^{n\ -\ 1}}]\)

Let, R is the radius of convergence. 

\(⇒ \ \frac{1}{R}\ =\ \mathop {\lim }\limits_{n \to {\rm{\infty}}}{|\frac{\frac{7^{n\ +\ 1}}{{(n\ +\ 1)}(5x\ -\ 1)^{^{n\ +\ 1\ -\ 1}}}}{\frac{7^n}{n(5x\ -\ 1)^{n\ -\ 1}}}|}\)

\(⇒ \ \frac{1}{R}\ =\ \mathop {\lim }\limits_{n \to {\rm{\infty}}}|\frac{7^{n + 1}(5x\ -\ 1)^{n-1}}{7^{n }(5x\ -\ 1)^{n}}|\)

\(⇒ \ \frac{1}{R}\ =\ \mathop {\lim }\limits_{n \to {\rm{\infty}}}|\frac{7n}{(n\ +\ 1)(5x\ -\ 1)}|\)

\(⇒ \ \frac{1}{R}\ =\ \mathop {\lim }\limits_{n \to {\rm{\infty}}}|\frac{7}{(1\ +\ \frac{1}{n})(5x\ -\ 1)}|\)

\(⇒ \ \frac{1}{R}\ =\ |\frac{7}{(5x\ -\ 1)}|\)      (∵ 1/n = 0)

⇒ R = \(\frac{1}{7}\)|5x - 1|

Given,

The given AP is 13, 11, 9……

Formula:

T_nt = a + (n – 1)d

a = first term

d = common term

Calculation:

a = 13

d = 11 – 13

d = (-2)

T₅₀ = 13 + (50 – 1) × (-2)

⇒ T₅₀ = 13 + 49 × (-2)

⇒ T₅₀ = 13 – 98

∴ T₅₀ = -85

Concept:

a + ar + ar2 + ar3 +….. 

Sum of the above infinite geometric series:

\(=\frac{a}{1-r}\)

Analysis:

Given:

1 + 2(a2 + 1) + 3(a2 + 1)2 + 4(a2 + 1)3 + ......

let x = (a2 + 1)

The series now becomes

S = 1 + 2x + 3x² + 4x³ + ......  ----(1)

By multiplying x on both sides we get

xS = x + 2x² + 3x³ + 4x⁴ + ...... ----(2)

Subtracting (1) and (2), we get

S(1 - x) = 1 + x + x² + x³ + ..... ---(3)

The right hand side of (3) forms infinite geometric series with a = 1, r = x

∴ S(1 - x) = \(\frac{1}{1-x}\)

\(\Rightarrow S = \frac{1}{(1-x)^2}\)

putting the value of x, we get

\(\Rightarrow S = \frac{1}{(1- a^2 - 1)^2}\)

\(\Rightarrow S = \frac{1}{a^4}\)

Formula used:

Sum of A.P. = n/2{first term + last term}

Calculation:

Number of terms = n = 12

⇒ S_n = 12/2{5 + 38}

⇒ S_n = 6{43}

Given:

The first five multiples of 3

Concept:

Multiples = A multiple is a number that can be divided by another number a certain number of time without a remainder

Calculation:

⇒ The first five multiple of 3 = (3 × 1), (3 × 2), (3 × 3), (3 × 4), (3 × 5) = 3, 6, 9, 12, and 15

⇒ The sum of the multiple = 3 + 6 + 9 + 12 + 15 = 45

∴ The required result will be 45.

Given:

Number are formed by using 6, 1, 2, 3

Calculation:

Number of ways to fill the first position = 4

Number of ways to fill second position = 3

Number of ways to fill third position = 2

Total number of ways = 4 × 3 × 2

⇒ Total number of ways = 24

∴ The total number of 3 digit number is 24.

Additional Information

If the repetition of number was allowed then we can take each digit in 4 different ways and so, the answer will be 4 × 4 × 4 = 64

Calculation:    

Total letters = 8

Letter 'E' is repeating  3 times  

Letter 'N' is repeating 2 times

⇒ Number of words can be formed = \(\frac{8!}{3! \times 2!}\)

⇒ Number of words can be formed = \(\frac{8 \times 7 \times 6 \times 5 \times 4\times 3\times 2\times 1 }{3 \times 2 \times 1\times 2 \times 1}\)

⇒ Number of words can be formed = 3360

∴ 3360 words can be formed.

The pattern followed here is –

Hence 16 will complete the series.

Shortcut Trick

Formula Used:

Average = (Sum of observations)/(Number of observations)

Last term = a + (n - 1)d

Calculation: 

The above series is in arithmetic progression so the middlemost term 8th term will be the average

⇒ 8^th term = 8 + (8 - 1) × 3 = 29

⇒ Sum of the series = 29 × 15 = 435

∴ The sum of the above series is 435  

Additional Information

We can avoid this above (29 × 15) multiplication by digit sum Method and option

The digit sum of 29 is  (2 + 9) ⇒ (11) ⇒ (2) and 15 is (1 + 5) = 6 

⇒ 2 × 6 = 12 ⇒ (1 + 2) ⇒ 3 

Now check the options whose digit sum will be 3 there is only option 2 whose Digit sum is 3 

∴ 435 is the right answer

Traditional Method: 

Given:

Arithmetic progression 8 + 11 + 14 + 17 upto 15 terms

Formula Used: 

Sum of arithmetic progression = n[2a + (n - 1)d]/2

Calculation:

Sum of 1st 15 terms = 15[2 × 8 + (15-1)3]/2

⇒ (15 × 58)/2

⇒ 435

∴ 435 is the right answer

Concept:

We have the factorial representation of \({^{n}C}_r\) as, \(\frac{{n\;!}}{{r\;! \times \left( {n - r} \right)!}}\).

Calculation:

\({^{12}C}_4\) = \(\frac{{12\;!}}{{4\;! \times \left( {12 - 4} \right)!}}\)

\({^{12}C}_4\) = \(\frac{{12\;!}}{{4\;! \times \left( {8} \right)!}}\)

\({^{12}C}_4\) = \(\frac{{12 \times11 \times 10 \times 9}}{{4 \times 3 \times 2 \times 1}}\)

\({^{12}C}_4\) = 495

Hence the value of \({^{12}C}_4\) is 495.

Concept:

Recurrence Relation:

A recurrence relation relates the nth term of a sequence to its predecessors. These relations are related to recursive algorithms.

Definition:

A recurrence relation for a sequence a0, a1, a2,.... is a formula (equation) that relates each term an to certain of its predecessors a0, a1, a2,...., an-1. The initial conditions for such a recurrence relation specify the values of a0, a1, a2,...., an-1. For example, the recursive formula for the sequence 3, 8, 13, 18, 23 is

a1 = 3, an = an-1 + 1, \(2\leq n<\infty\)

Calculation:

Given:

The recurrence relation , T(n) = T(n - 1)+ 2

If n = 1  then T(n) = T(n-1)+ 2 = T(1) = T(1-1)+ 2 = T(0) + 2 =5 + 2 = 7   / Value of T(0) given in Question

If n= 2  then T(n) = T(n-1)+ 2 = T(1) = T(2-1)+ 2 = T(1) + 2 =7 + 2 = 9   / Value of T(1) is 7 

If n= 3  then T(n) = T(n-1)+ 2 = T(1) = T(3-1)+ 2 = T(2) + 2 =9 + 2 = 11   / Value of T(2) is 9

Therefore, above pattern can be written in the form of

T(n) = 2n+ 5 

If n= 1 then T(n) = 2n+ 5 = T(1) = 2(1)+ 5 = T(1) =7 

Therefore Option 3 is the correct Answer