Permutations and Combinations Test 27

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Permutations and Combinations Test 27

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Weekly Quiz Competition

Question 1
1 / -0

If the letters of the word "VARUN" are written in all possible ways and then are arranged as in a dictionary, then the rank of the word VARUN is?

A
$$98$$

B
$$99$$

C
$$100$$

D
$$101$$

Solution

$${\textbf{Step 1}}\;:\;{\mathbf{Finding}}\;{\mathbf{number}}\;{\mathbf{of}}\;{\mathbf{words}}\;{\mathbf{starting}}\;{\mathbf{with}}\;{\mathbf{the}}\;{\mathbf{alphabets}}\;{\mathbf{in}}\;{\mathbf{the}}\;{\mathbf{given}}\;{\mathbf{word}}$$

$${\text{No}}.{\text{ of words beginning with A }} = 4! = 4 \times 3 \times 2 = 24$$

$${\text{No}}.{\text{ of words beginning with N }} = 4! = 4 \times 3 \times 2 = 24$$

$${\text{No}}.{\text{ of words beginning with R }} = 4! = 4 \times 3 \times 2 = 24$$

$${\text{No}}.{\text{ of words beginning with U }} = 4! = 4 \times 3 \times 2 = 24$$

$${\text{So}},{\text{ in total }}96{\text{ words will be formed while beginning with letter A}},{\text{ N}},{\text{ R and U}}.$$

$${\textbf{Step 2}}\;\;:\;{\mathbf{Finding}}\;{\mathbf{the}}\;{\mathbf{order}}\;{\mathbf{of}}\;{\mathbf{the}}\;{\mathbf{word}}$$

$${\text{Order of }} 97{\text { th word }} - {\text{ VANRU}}$$

$${\text{Order of }}98{\text{th word }} - {\text{ VANUR}}$$

$${\text{Order of }}99{\text{th word }} - {\text{ VARNU}}$$

$${\text{Order of }}100{\text{th word}} - {\text{ VARUN}}.$$

$${\mathbf{Therefore}},\;{\mathbf{rank}}\;{\mathbf{of}}\;{\mathbf{word}}\;'{\mathbf{VARUN}}'\;{\textbf{is 100. Option C is correct.}}$$
Question 2
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Garlands are formed using 6 red roses and 6 yellow roses of different sizes. The number of arrangements in garland which have red roses and yellow roses come alternately is

A
$$5!\times 6!$$

B
$$6! \times 6!$$

C
$$\dfrac{5!}{2!}\times 6!$$

D
$$2(6!\times 6!)$$

Solution
Question 3
1 / -0

The value of $$\sum\limits_{r = 0}^{10} {\left( {\matrix{ {10} \cr r \cr } } \right)} \left( {\matrix{ {15} \cr {14 - r} \cr } } \right)$$ is equal to

A
$${}^{25}{C_{12}}$$

B
$${}^{25}{C_{15}}$$

C
$${}^{25}{C_{10}}$$

D
$${}^{25}{C_{11}}$$

Solution
Question 4
1 / -0

If $$f(x)=1-x+x^2-x^3+....-x^{15}+x^{16}+x^{17}$$, then the coefficient of $$x^2$$ in $$f(x-1)$$ is?

A
$$826$$

B
$$816$$

C
$$822$$

D
None of these

Solution

$$f\left( x \right) = 1 - \left( {x - 1} \right) + {\left( {x - 1} \right)^2} - {\left( {x - 1} \right)^3} + {\left( {x - 1} \right)^4}.... - {\left( {x - 1} \right)^{15}} + {\left( {x - 1} \right)^{16}} + {\left( {x - 1} \right)^{17}}$$
The coefficient of $${x^2}$$ by using binomial expansion
$$^2{c_0}{x^2}{ + ^3}{c_1}{x^2}{ + ^4}{c_2}{x^2}{ + ^5}{c_3}{x^2}{ + ^6}{c_4}{x^2}{ + ^7}{c_5}{x^2}{ + ^8}{c_6}{x^2}{ + ^9}{c_7}{x^2}$$
$${ + ^{10}}{c_8}{x^2}{ - ^{11}}{c_9}{x^2}{ + ^{12}}{c_{10}}{x^2}{ - ^{13}}{c_{11}}{x^2}{ + ^{14}}{c_{12}}{x^2}{ - ^{15}}{c_{13}}{x^2}{ + ^{16}}{c_{14}}{x^2}{ - ^{17}}{c_{15}}{x^2}$$
$$ \Rightarrow {x^2}\left( {1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 + 45 + 55 + 66 + 78 + 91 + 105 + 120 + 136} \right)$$
= 372 + 444 = 816
Question 5
1 / -0

The product of five consecutive numbers is always divisible by ?

A
$$60$$

B
$$12$$

C
$$120$$

D
$$72$$

Solution

Solution:
The product $$(n)$$ of any $$5$$ consecutive numbers will always be divisible by $$120$$ and its factors.
$$n(n+1)(n+2)(n+3)(n+4)$$
$$5\times 4\times 3\times 2=120$$
Answer C:
Question 6
1 / -0

A person tries to form as many different parties as he can, out of his $$20$$ friends. Each party should consist of the same number. How many friends should be invited at a time? In how many of these parties would the same friends be found?

A
92378

B
92364

C
92376

D
92391

Solution

Let the person invite r number of friends at a time. Then, the number of parties is $$^{20}C_r$$, which is maximum when $$r=10$$. If a particular friend will be found in x parties, then x is the number of combinations out of $$20$$ in which this particular friend must be included. Therefore, we have to select $$9$$ more from $$19$$ remaining friends. Hence, $$x=$$ $$^{19}C_9$$=92378.
Question 7
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Let $$(1+x)^n=C_0+C_1x+C_2x^2+....+C_nX^n$$.(where $$C_r=$$ $$^{n}C_r$$). On the basis of information, answer the following question.
$$2(C_2)-4(C_4)+6(C_6)$$_______ is?

A
$$(\sqrt{2})^{\dfrac{(n-1)}{2}}$$

B
$$2^{\dfrac{(n-1)}{2}}.n.\sin\left(\dfrac{(n-1)\pi}{4}\right)$$

C
$$\sin \dfrac{n\pi}{4}$$

D
$$2^{\dfrac{(n-1)}{2}}.n.\sin\dfrac{n\pi}{4}$$
Question 8
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Let $$5 < n_1 < n_2 < n_3 < n_4$$ be integers such that $$n_1+n_2+n_3+n_4=35$$. The number of such distinct arrangements $$(n_1, n_2, n_3, n_4)$$.

A
$$^{38}C_3$$

B
$$^8C_3$$

C
$$5$$

D
$$6$$

Solution

$${ n }_{ 1 }+{ n }_{ 2 }+{ n }_{ 3 }+.....+{ n }_{ k }=R$$
for this arrangement,
No. of arrangement or distinct arrangements are $${ R+K-1 }_{ { C }_{ K-1 } }$$
So, for $${ n }_{ 1 }+{ n }_{ 2 }+{ n }_{ 3 }+{ n }_{ 4 }=35$$
No. of arrangements $$={ 35+4-1 }_{ { C }_{ 4-1 } }={ 38 }_{ { C }_{ 3 } }$$
Question 9
1 / -0

The value of $${ _{ }^{ 13 }{ C } }_{ 2 }+{ _{ }^{ 13 }{ C } }_{ 3 }+{ _{ }^{ 13 }{ C } }_{ 4 }+....+{ _{ }^{ 13 }{ C } }_{ 13 }$$ is

A
$${2}^{13}-13$$

B
$${2}^{13}-14$$

C
an odd number $$\ne {2}^{13}-12$$

D
an even number $$\ne {2}^{13}-14$$

Solution

$${\textbf{Step -1: Simplifying the problem.}}$$
  $${}^{13}{C_2} + {}^{13}{C_3} + ...... + {}^{13}{C_{13}}.$$
$$ = {}^{13}{C_0} + {}^{13}{C_1} + {}^{13}{C_2} + {}^{13}{C_3} + ...... + {}^{13}{C_{13}} - {}^{13}{C_0} - {}^{13}{C_1}.$$
$$ = \left( {{}^{13}{C_0} + {}^{13}{C_1} + {}^{13}{C_2} + ..... + {}^{13}{C_{13}}} \right) - \left( {{}^{13}{C_0} + {}^{13}{C_1}} \right){\text{   }} \to \left( {\text{i}} \right).$$
$${\textbf{Step -2: Deduce the value of }}\mathbf{\left( {{}^{13}{C_0} + {}^{13}{C_1} + {}^{13}{C_2} + ..... + {}^{13}{C_{13}}} \right)}{\textbf{ using summation formula}}{\textbf{. }}$$
$$\sum\limits_{i = 0}^n {{}^n{C_i}}  = {2^n}{\text{   }} \to \left( {{\text{ii}}} \right)$$
         $$ \Rightarrow \left( {{}^{13}{C_0} + {}^{13}{C_1} + {}^{13}{C_2} + ..... + {}^{13}{C_{13}}} \right) = \sum\limits_{i = 0}^{13} {{}^{13}{C_i}.} $$
  $$ \Rightarrow \sum\limits_{i = 0}^{13} {{}^{13}{C_i} = {2^{13}}({\text{By using (}}ii{\text{)}})} .$$
   $${\text{then, using (i) and (ii) we get}},$$
   $$\therefore {2^{13}} - \left( {{}^{13}{C_0} + {}^{13}{C_1}} \right).$$
$$ = {2^{13}} - \left( {\dfrac{{13!}}{{13!0!}} + \dfrac{{13!}}{{12!1!}}} \right).$$
$$ = {2^{13}} - \left( {1 + 13} \right).$$
             $$ = {2^{13}} - 14.$$
$${\textbf{Hence , the correct answer is (B) }}\mathbf{{2^{13}} - 14.}$$
Question 10
1 / -0

In an examination there are three multiple choice questions and each question has $$4$$ choices. Number of ways in which a student can fail to get all answer correct is?

A
$$11$$

B
$$12$$

C
$$27$$

D
$$63$$

Solution

There are 4 choices for each question. So, there
are 4 ways to answer a question.

Number of ways to answer 3 questions

$$=4\times 4\times 4$$
$$\downarrow $$ $$\downarrow $$ $$\downarrow $$
$$1^{st}$$ question $$2^{nd}$$ $$3^{rd}$$
$$= 64$$

Out of 64 ways there is only one way which has
all the answers correct.

So, Number of ways in which a student can fail to get all answers
correct $$= 64-1$$

$$= 63$$

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Re-Attempt Weekly Quiz Competition

Permutations and Combinations Test 27

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Permutations and Combinations Test 27

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Question 1

$$98$$

$$99$$

$$100$$

$$101$$

Solution

Question 2

$$5!\times 6!$$

$$6! \times 6!$$

$$\dfrac{5!}{2!}\times 6!$$

$$2(6!\times 6!)$$

Solution

Question 3

$${}^{25}{C_{12}}$$

$${}^{25}{C_{15}}$$

$${}^{25}{C_{10}}$$

$${}^{25}{C_{11}}$$

Solution

Question 4

$$826$$

$$816$$

$$822$$

None of these

Solution

Question 5

$$60$$

$$12$$

$$120$$

$$72$$

Solution

Question 6

92378

92364

92376

92391

Solution

Question 7

$$(\sqrt{2})^{\dfrac{(n-1)}{2}}$$

$$2^{\dfrac{(n-1)}{2}}.n.\sin\left(\dfrac{(n-1)\pi}{4}\right)$$

$$\sin \dfrac{n\pi}{4}$$

$$2^{\dfrac{(n-1)}{2}}.n.\sin\dfrac{n\pi}{4}$$

Question 8

$$^{38}C_3$$

$$^8C_3$$

$$5$$

$$6$$

Solution

Question 9

$${2}^{13}-13$$

$${2}^{13}-14$$

an odd number $$\ne {2}^{13}-12$$

an even number $$\ne {2}^{13}-14$$

Solution

Question 10

$$11$$

$$12$$

$$27$$

$$63$$

Solution

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