MCQ Questions for Class 11 Engineering Maths Permutations And Combinations Quiz 1

$\displaystyle { S }_{ { n } }=\sum _{ { r }=0 }^{ { n } } \frac { 1 }{ ^{ n }{ { C }_{ r } } }$ and

$\displaystyle { t }_{ { n } }=\sum _{ { r }=0 }^{ { n } } \frac { { r } }{ ^{ n }{ { C }_{ r } } }$ , then

$\displaystyle \frac { { t }_{ { n } } }{ { s }_{ { n } } } =$

0%
$\displaystyle \frac{1}{2}{n}$
0%
$\displaystyle \frac{1}{2}n-1$
0%
${n}-1$
0%
$\displaystyle \frac{2{n}-1}{2}$

Explanation

$\displaystyle{ S }_{ n }=\sum _{ r=0 }^{ n }{ \frac { 1 }{ _{ }^{ n }{ { C }_{ r } } } } =\sum _{ r=0 }^{ n }{ \frac { 1 }{ _{ }^{ n }{ { C }_{ n-r } } } } \left( \because _ { } ^{ n }{ { C }_{ r } }=^{ n }{ { C }_{ n-r } } \right)$

$\displaystyle{ t }_{ n }=\sum _{ r=0 }^{ n }{ \frac { r }{ _{ }^{ n }{ { C }_{ n-r } } } }$

$\displaystyle\therefore n{ S }_{ n }=\sum _{ r=0 }^{ n }{ \left[ \frac { n-r }{ ^{ n }{ { C }_{ n-r } } } +\frac { r }{ ^{ n }{ { C }_{ n-r } } } \right] }$

$\displaystyle=\sum _{ r=0 }^{ n }{ \frac { n-r }{ ^{ n }{ { C }_{ n-r } } } } +\sum _{ r=0 }^{ n }{ \frac { r }{ _{ }^{ n }{ { C }_{ r } } } }$

$\displaystyle=\left( \frac { n }{ _{ }^{ n }{ { C }_{ n } } } +\frac { n-1 }{ ^{ n }{ { C }_{ n-1 } } } +...+\frac { 0 }{ _{ }^{ n }{ { C }_{ n } } } \right) +\sum _{ r=0 }^{ n }{ \frac { r }{ _{ }^{ n }{ { C }_{ r } } } }$

$\displaystyle\Rightarrow n{ S }_{ n }={ t }_{ n }+{ t }_{ n }=2{ t }_{ n }$

$\displaystyle\Rightarrow \frac { { t }_{ n } }{ { S }_{ n } } =\frac { n }{ 2 }$

The value of

$^{50}C_{4}+ \sum _{r=1}^{6}\ ^{56-r}C_{3}$ is

0%
$^{55}C_{4}$
0%
$^{55}C_{3}$
0%
$^{56}C_{3}$
0%
$^{56}C_{4}$

Explanation

$^{50}C_{4}+\sum _{r=1}^{6}\ ^{56-r}C_{3}$

$\Rightarrow ^{50}C_{4}+[^{55}C_{3}+^{54}C_{3}+^{53}C_{3}+^{52}C_{3}+^{51}C_{3}+^{50}C_{3}]$

$=(^{50}C_{4}+^{50}C_{3})+^{51}C_{3}+^{52}C_{3}+^{53}C_{3}+^{54}C_{3}+^{55}C_{3}$ ........................ As

$(^{n}C_{r}+^{n}C_{r-1}=^{n+1}C_{r})$

$=(^{51}C_{4}+^{51}C_{3})+^{52}C_{3}+^{53}C_{3}+^{54}C_{3}+^{55}C_{3}$

$=\ ^{52}C_{4}+^{52}C_{3}+^{53}C_{3}+^{54}C_{3}+^{55}C_{3}$

$=\ ^{53}C_{4}+^{53}C_{3}+^{54}C_{3}+^{55}C_{3}$

$=\ ^{54}C_{4}+^{54}C_{3}+^{55}C_{3}$

$\Rightarrow ^{55}C_{4}+^{55}C_{3}=^{56}C_{4}$

If all the words, with or without meaning, are written using the letters of the word QUEEN and are arranged as in English dictionary, then the position of the word QUEEN is.

0%
$45^{th}$
0%
$46^{th}$
0%
$44^{th}$
0%
$47^{th}$

Explanation

$\textbf{Step -1:Possible combination of words in each set.}$

$\text{Putting the word QUEEN in alphabetical order, we get, }$

$\text{EENQU}$

$\text{If a word with or without meaning starts with E, then}$

$\text{E _ _ _ _, the four blanks can be filled in 4! ways.}$

$\text{If a word with or without meaning starts with N, then}$

$\text{N _ _ _ _, the four blanks can be filled in }\dfrac{4!}{2!} \text{ ways as E is repeating twice.}$

$\text{If a word with or without meaning starts with QE, then}$

$\text{Q E _ _ _, the three blanks can be filled in 3! ways.}$

$\text{If a word with or without meaning starts with QN, then}$

$\text{Q N _ _ _, the three blanks can be filled in }\dfrac{3!}{2!} \text{ ways as E is repeating twice.}$

$\text{The word QUEEN comes soon after the above series, hence 1!}$

$\textbf{Step -2:Calculating the position of given word.}$

$\text{The position of the word QUEEN = }4! + \dfrac{4!}{2!} + 3! + \dfrac{3!}{2!} + 1!$

$= 24 + \dfrac{24}{2} + 6 + \dfrac{6}{2} + 1$

$= 24 + 12 + 6 + 3 + 1$

$= 46$

$\textbf{Hence, the correct option is B.}$

Statement-l:

$\displaystyle \sum_{r=0}^{n}(r+1)^{n}C_{r}=(n+2)2^{n-1}$
Statement-2:

$\displaystyle \sum_{r=0}^{n}(r+1)^{n}C_{r}x^{r}=(1+x)^{n}+nx(1+x)^{n-1}$

0%
Statement-1 is true, Statement-2 is true; Statement -2 is not a correct explanation for Statement-1.
0%
Statement-1 is true, Statement-2 is false.
0%
Statement-1 is false, Statement-2 is true.
0%
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.

Explanation

$\displaystyle \sum_{r=0}^{n}(r+1)^nC_{r}x^r=\sum_{r=0}^{n}r.C_{r}x^{r}+\sum_{r=0}^{n}C_{r}.x=nx \displaystyle \sum_{r=1}^{n}C_{r-1}x+\sum_{r=0}^{n}C_{r}x=nx(1+x)^{n-1}+(1+x)^{n}$ ... (i)

$\Rightarrow$ Statement-2 is true.
Putting

$x=1$ in (i), we get

$\displaystyle \sum_{r=0}^{n}(r+1)n_{C_{r}=}(n+2).2^{n-1}\Rightarrow$ Statement-l is also true and followed by I.

The total number of 4-digit numbers in which the digits are descending order is

0%
${ _{ }^{ 10 }{ C } }_{ 4 }\times 4!$
0%
${ _{ }^{ 10 }{ C } }_{ 4 }$
0%
$\cfrac { 10! }{ 4! }$
0%
None of these

Explanation

Total number of arrangements of

$10$ digits

$0,1,2,....9$ by taking

$4$ at at time

${ _{ }^{ 10 }{ C } }_{ 4 }\times 4!\quad$
We observe that in every arrangement of

$4$ selected digits there is just one arrangement in which the digits are in descending order

$\therefore$ Required number of

$4$ -digit numbers

$=\cfrac { { _{ }^{ 10 }{ C } }_{ 4 }\times 4! }{ 4! } ={ _{ }^{ 10 }{ C } }_{ 4 }$

An automobile dealer provides motorcycles and scooters in three body patterns and

$4$ different colors each. The number of choices open to a customer is

0%
$^5C_3$
0%
$^4C_3$
0%
$4\times 3$
0%
$4\times 3\times 2$

Explanation

No. of vehicle options (Motorcycle, scooter)

$=2$

No. of body patterns

$=3$

No. of different colors

$=4$

No. of options for customer=

$\underset{2}{\underset{\downarrow}{\underset{|-------------------|} {No. \, of \, vehicle\, options}}} {\underset{X} {\underset{\downarrow}{AND}}} {\underset{3}{\underset{\downarrow}{\underset {|---------------------|} {No.\, of \, body\, pattern}}}} {\underset {X} {\underset {\downarrow} {AND}}} {\underset{4}{\underset {\downarrow} {\underset{|------------------------|}{No.\, of \, colors}}}}$

$= 4 \times 3 \times 2$ .

There are

$'m'$ copies each of

$'n'$ different books in a university library. The number of ways in which one or more than one book can be selected is

0%
$m^n - 1$
0%
$(m+1)^n - 1$
0%
$(m+1)^n - m^n$
0%
$(m+1)^n - m$

Explanation

There are

$m$ copies each of

$n$ different books.
So, we can choose any of the

$m$ books of each copy or we don't choose any of them.
So, we have

$\left( m+1 \right)$ choices for each of

$n$ different books.
But we have to select at least one book. So we will subtract a case when none of the books got selected.
Hence, no. of ways of selecting one or more books is

$\left( \begin{matrix} \left( m+1 \right) \\ 1 \end{matrix} \right) \left( \begin{matrix} \left( m+1 \right) \\ 1 \end{matrix} \right) \left( \begin{matrix} \left( m+1 \right) \\ 1 \end{matrix} \right) .......n$ times

$-1={ \left( m+1 \right) }^{ n }-1$

In a crossword puzzle,

$20$ words are to be guessed of which

$8$ words have each an alternative solution also. The number of possible solutions will be

0%
$^{20}P_8$
0%
$^{20}C_8$
0%
$512$
0%
$256$

Explanation

$8$ out of the

$20$ words have an alternative solution.

$12$ words are fixed, so they won't affect the number of ways.

So now we find the number of ways in which the remaining

$8$ words can be guessed.

Now every word have

$2$ solutions,

so there are

$2$ ways to fill each of these words, and for

$8$ such words,

it can be done in

$2\times 2\times 2\times 2\times 2\times 2\times 2\times 2={ 2 }^{ 8 }=256$ ways.

$^xC_7 - ^xC_5 = 0$ , then

$x =$

0%
7
0%
5
0%
12
0%
10

Explanation

Given,

$^xC_7 - ^xC_5 = 0$

$\Rightarrow ^xC_7 = ^xC_5$
Now we know,

$^nC_x = ^nC_y$

$\Leftrightarrow x+y=n$

$\therefore x = 7+5=12$

$15$ buses operate between Hyderabad and Tirupathi.The number of ways can a man go to Tirupathi from Hyderabad by a bus and return by a different bus is

0%
$15$
0%
$150$
0%
$210$
0%
$225$

Explanation

While going ,number of ways to choose

$1$ bus out of

$15$ is

$^{15}C_{1}$ .
While return trip, in order to come with different bus, number of buses left

$=14$ .
Number of ways to choose

$1$ bus for return trip

$=$

$^{14}C_{1}$
So, the required number of ways for going and return

$=$

$^{15}C_{1} \times ^{14}C_{1}=210$

The number of nine digit numbers that can be formed with different digits is

0%
$9. 8!$
0%
$8 . 9!$
0%
$9. 9!$
0%
$10!$

Explanation

The first digit is one of

$(1, 2, 3, 4, 5, 6, 7, 8, 9)$ (i.e., all expect

$0$ ).
So, the no. of ways

$= 9$
The remaining

$8$ digits are to be taken from the remaining

$9$ numbers.
So, the no. of ways

$= ^9P_8$
So, total no. of ways

$= 9 \times ^9P_8$

$= 9 \times 9!$

There are

$5$ doors to a lecture hall. The number of ways that a student can enter the hall and leave it by a different door is

0%
$20$
0%
$16$
0%
$19$
0%
$21$

Explanation

A student has 5 different doors to enter,
Every door a student enters through, there are 4 more doors to leave through.

So,

$\text{Total no. of ways}\,=5 \times 4 = 20$

There are

44 candidates for a Natural science scholarship,

22 for a Classical and

66 for a Mathematical scholarship,then find the no. of ways one of these scholarship can be awarded is,

0%
$6$
0%
$10$
0%
$48$
0%
$12$

Explanation

Natural science scholarship can be awarded to anyone of the four candidates. So, there are

$4$ ways of awarding natural science scholarships.

Similarly, mathematical and classical scholarships can be awarded in

$2$ and

$6$ ways respectively.

By the fundamental principle of addition, number of ways of awarding one of the three scholarships is

$=4+2+6=12$

Hence option-

$D$

Choose the correct option for the following.

$n!=n(n-1)(n-2).....3.2.1$

0%
True
0%
False
0%
Ambiguous
0%
Data insufficient

Explanation

Factorial : the product of an integer and all integer less than that

$\therefore n!=n\times (n-1)\times (n-2)............3\times 2\times 1$

$\therefore$ The given statement

$n!=n\times (n-1)\times (n-2)............3\times 2\times 1$ is True

Five persons A, B, C, D and E occupy seats in a row such that A and B sit next to each other. In how many possible ways can these five people sit?

0%
24
0%
48
0%
72
0%
56

Explanation

(B)

$4!\times 2$ ways. i.e,

$24\times 2=48$ ways

Factorial of negative numbers is always greater than 1.

0%
True
0%
False
0%
Either
0%
Neither

Ten different letter of an alphabet are given. Words with five letters are formed from these given letters. Then the number of words which have at least one letter repeated is:

0%
19670
0%
39758
0%
69760
0%
99748

Value of 0! is always 1.

0%
True
0%
False
0%
Either
0%
Neither

Explanation

we know

$1!=1$

Also

$n!=n\times (n-1)\times (n-2)........3\times 2\times 1\\ n!=n\times (n-1)!\\ 1!=1(1-1)!\\ 1=1(0)!\\ 0!=1$

$0!$ is always

$1$

Hence its true that

$0!$ is always

$1$

Each combination corresponds to many permutations.

0%
True
0%
False
0%
Either
0%
Neither

The value of

$\displaystyle ^{ 6 }C_{ 4 }$ is

0%
$6$
0%
$9$
0%
$15$
0%
$240$

Explanation

$\displaystyle ^{ 6 }C_{ 4 }=^{ 6 }C_{ 6-4 }=^{ 6 }C_{ 2 }=\frac { 6\times 5 }{ 2\times 1 } =15$

A batsman can score

$0,1,2,3,4$ or

$6$ runs from a ball. The number of different sequences in which he can score exactly

$30$ runs in an over of six balls is:

0%
$4$
0%
$72$
0%
$56$
0%
$71$

Explanation

Possible cases of scoring

$30$ runs:

Case I:

$\{6,6,6,6,6,0\}$

Total sequence:

$\dfrac{6!}{5!}=6$

Case II:

$\{6,6,6,6,4,2\}$

Total sequence:

$\dfrac{6!}{4!} = 30$

Case III:

$\{6,6,6,6,3,3\}$

Total sequence:

$\dfrac{6!}{2!4!} = 15$

Case IV:

$\{6,6,6,4,4,4\}$

Total sequence:

$\dfrac{6!}{3!3!} = 20$

Hence total number of sequence is:

$6+30+15+20 = 71$

How many combinations of two-digit numbers having 8 can be made from the following numbers?
8, 5, 2, 1, 7, 6

0%
$9$
0%
$10$
0%
$11$
0%
$12$

On the eve of Diwali festival, a group of 12 friends greeted every other friend by sending greeting cards. Find the number of cards purchased by the group.

0%
156
0%
144
0%
132
0%
72
0%
66

A group consists of 4 couples in which each of the 4 persons have one wife each. In how many ways could they be arranged in a straight line such that the men and women occupy alternate positions?

0%
1152
0%
1278
0%
1296
0%
1176

Explanation

(A) Required no of ways

$=2\times 4!\times 4!$

$=1152$

The greatest number that can be formed by the digits

$7,0,9,8,6,3$

0%
$9,87,360$
0%
$9,87,063$
0%
$9,87,630$
0%
$9,87,603$

Explanation

The greatest number that can be formed by the digits

$7,0,9,8,6,3$ is

$9 8 7 6 3 0$

To achieve this arrange the given numbers in descending order.

In the series given below. count the number of 9s, each of which Is not immediately preceded by 5 but is immediately followed by either 2 orHow many such 9s are there?
1 9 2 6 5 9 3 8 3 9 3 2 5 9 2 9 3 4 8 2 6 9 8

0%
One
0%
Three
0%
Five
0%
Six

How many numbers of four digits can be formed from the digits 0, 1, 2, 3, and 4?

0%
48
0%
64
0%
96
0%
100

Explanation

The first box can be filled in four ways, because we cannot put 0 in the first box. The second box can also be filled in four ways, because we cannot put 0. The third box can be filled in three ways and the fourth in two ways. Therefore,
Total numbers =

$\displaystyle 4\times 4\times 3\times 2=96$
1294467_375391_ans_16169efc886947cca81ec3eda1c1405e.PNG

How many numbers amongst the numbers 9 to 54 are there which are exactly divisible by 9 but not by 3?

0%
8
0%
6
0%
5
0%
Nil

Arrange the given words in the sequence in which they occur in the dictionary and then choose the correct sequence.

PagePaganPalisadePageant Palate

0%
$1,4,2,3,5$
0%
$2,4,1,3,5$
0%
$2,1,4,5,3$
0%
$1,4,2,5,3$

In a class there are 18 boys who are over 160 cm tall If these constitute three-fourths of the boys and the total number of boys is tow-third of the total number of students in the class what is the number of girls in the class?

0%
6
0%
12
0%
18
0%
24

Explanation

Given in class there are 18 boys who over 160 cm tall and these are three- fourths of total boys

Then total boys in class=

$18\div \frac{3}{4}= 18\times \frac{4}{3}=24$

And total no of student in class=

$24\times \frac{3}{2}=36$

Then number of girls=36-24=12

Out of 100 students 50 fail in English and 30 in Maths. If 12 students fail in both English and Maths, then the number of students passing both the subjects is

0%
26
0%
28
0%
30
0%
32

At the end of a business conference, the ten people present all shake hands with each other once. How many handshakes will there be altogether?

0%
20
0%
45
0%
55
0%
90

A group of 1200 persons consisting of captains and soldiers is travelling in a train. For every 15 soldiers there is one captain. The number of captains in the group is:

0%
85
0%
80
0%
75
0%
70

Explanation

Out of 16 men, there is a captain.
Number of captains in 1200 men =

$\displaystyle \\ 1200\div 16=75$ .

A car driver knows four different routes from Delhi to Amritsar. From Amritsar to Pathankot, he knows three different routes and from Pathankot to Jammu he knows two different routes. How many routes does he know from Delhi to Jammu?

0%
4
0%
8
0%
12
0%
24
0%
36

Explanation

The car driver can reach Amritsar in 4 ways.From each of these four ways, he can reach Pathankot in 3 different ways and so he can reach from Delhi to Pathankot in

$\displaystyle \left( 4\times 3 \right)$ i.e 12 ways. Again from Pathankot to Jammu, there are 2 ways. Hence he can reach Jammu from Delhi in

$\displaystyle \left( 12\times 2 \right)$ i.e. in 24 ways.

In a chess tournament each of six players will play every other player exactly once. How many matches will be played during the tournament?

0%
36
0%
30
0%
15
0%
12

Explanation

First player can play 5 matches with other five players. Second player can play 4 matches with other four players and proceeding this way, the fifth player will play only one match with sixth player.

$\displaystyle \therefore$ Total number of matches played = 5 +4 + 3 + 2 + 1 = 15.
Short cut : Number of matches played by n players =

$\displaystyle \frac { n\left( n+1 \right) }{ 2 }$

Amy and Adam are making boxes of truffles to give out as wedding favors. They have an unlimited supply of 5 different types of truffles. If each box holds 2 truffles of different types, how many different boxes can they make?

0%
$12$
0%
$10$
0%
$15$
0%
$20$

Explanation

10 boxes
In every combination, 2 types of truffles will be in the box, and 3 types of truffles will not. Therefore, this problem is a question about the number of anagrams that can be made from the "word" YYNNN:

$\dfrac{5!}{2!3!} = \dfrac{5\times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1 \times 2 \times 1} = 5 \times 2 = 10$

Let the coefficient of

$6^{th}$ term of an expansion be

$a$ and

$b$ be the power.
Expansion:

$\left (\dfrac{4x}{5}- \dfrac 5{2x}\right)^{9}$
Find

$a \times b$ .

0%
$4800$
0%
$5040$
0%
$-2700$
0%
$-14400$

Explanation

$(r+1)^{th}$ term is

$t_{r+1}= ^nC_r a^{n-r} b^r$
So, for

$6^{th}$ term,

$r=5$ .

$\therefore t_{6}=t_{5+1}=^{9}C_5 (-1)^5(\dfrac{4x}{5})^{9-5} (\dfrac 5{2x})^5$

$\therefore t_{6}=\dfrac {-5040}{x}$
So,

$a=-5040, b=-1$
Ans-Option

$B$ .

A garrison of '

$n$ ' men had enough food to last for

$30$ days. After

$10$ days,

$50$ more men joined them. If the food now lasted for

$16$ days, what is the value of

$n$ ?

0%
$200$
0%
$240$
0%
$280$
0%
$320$

Explanation

After

$10$ days, the food for n men is there for

$20$ days.
This food can be eaten by

$(n + 50)$ men in

$16$ days.

$\therefore 20n = 16(n + 50)$

$\therefore n = 200$

Based on this information answer the questions given below.
(i)

$\displaystyle ^{ n }{ C }_{ p }= r!^{ n }{ C }_{ r }$
(ii)

$\displaystyle ^{ n }{ C }_{ r }+^{ n }{ C }_{ r-1 }=^{ n+1 }{ C }_{ r }$

What is the value of

$\displaystyle ^{ 8 }{ C }_{ 4 }+^{ 8 }{ C }_{ 3 }$ ?

0%
$\displaystyle ^{ 8 }{ c }_{ 5 }$
0%
$63$
0%
$35$
0%
$\displaystyle ^{ 9 }{ c }_{ 4 }$

Explanation

$^{n}C_{r}+^{n}C_{r-1} = ^{n+1}C_{r}$

$\Rightarrow ^{8}C_{4}+^{8}C_{3} = \boxed{^{9}C_{4}}$

1108234_518590_ans_1e3a51edefcb4e53be5ef87a3f12d81e.jpg

A bag contains Rs.

$112$ in the form of

$1$ -rupee,

$50$ -paise and

$10$ -paise coins in the ratio

$3 : 8 : 10$ . What is the number of

$50$ -paise coins?

0%
$112$
0%
$128$
0%
$96$
0%
$24$

Explanation

Here bag contains Rs.

$112$ in the form of

$1$ -rupee,

$50$ -paise and

$10$ -paise.

Coin ratio is

$3 : 8 : 10$ .
Value ratio

$= 3 \times 1: 8\times \dfrac {1}{2} : 10 \times \dfrac {1}{10}$

$= 3 : 4 : 1$
Number of

$50$ paise coins

$= \dfrac {4}{8} \times$ Rs.

$112 =$ Rs.

$56$
Therefore, number of

$50$ paise coins

$= 2\times 56 = 112$ .

Let the coefficient of

$10^{th}$ term of an expansion be

$a$ and

$b$ be the power.
Expansion:

$\left (2x^2+ \dfrac 1x\right)^{12}$
Find

$a \times b$ .

0%
$-5280$
0%
$4200$
0%
$-3460$
0%
$7360$

Explanation

$(r+1)^{th}$ term is

$t_{r+1}= ^nC_r a^{n-r} b^r$
So, for

$10^{th}$ term,

$r=9$ .

$\therefore t_{10}=t_{9+1}=^{12}C_9 (2x^2)^{12-9} (\dfrac 1x)^9$

$\therefore t_{10}=\dfrac {1760}{x^3}$
So,

$a=1760, b=3$
Ans-Option

$A$ .

Let the coefficient of

$5^{th}$ term from the end of an expansion be

$a$ and

$b$ be the power.
Expansion:

$\left (\dfrac{x^3}{2}- \dfrac 2{x^2}\right)^{9}$
Find

$a \times b$ .

0%
$424$
0%
$-252$
0%
$-504$
0%
$335$

Explanation

$5^{th}$ from the end is the

$6^{th}$ term.

$(r+1)^{th}$ term is

$t_{r+1}= ^nC_r a^{n-r} b^r$
So, for

$6^{th}$ term,

$r=5$ .

$\therefore t_{6}=t_{5+1}=^{9}C_4 (-1)^5(\dfrac{x^3}{2})^{9-5} (\dfrac 2{x^2})^5$

$\therefore t_{6}=\dfrac {-252}{x^2}$
So,

$a=-252, b=2$
Ans-Option

$C$ .

In how many ways can a group of

$5$ men and

$2$ women be made out of a total of

$7$ men and

$3$ women

0%
$90$
0%
$126$
0%
$63$
0%
$43$

Explanation

Out of

$7$ men and

$3$ women , a group of

$5$ men and

$2$ women can be made by

$^7C_5 \times ^3C_2=63$

$^{20}C_{r+2} = ^{20}C_{2r-3}$ , then find

$^{12}C_r$

0%
$642$
0%
$780$
0%
$792$
0%
$256$

Explanation

$r+2 =20-2r+3$

$\Longrightarrow r+2 +2r-3 =20$

$\Longrightarrow 3r-1 =20$

$r=7$

$^{12}C_7$

$=792$

The number of rectangles that you can find on a chess board is :

0%
$1442$
0%
$1296$
0%
$1256$
0%
None of these

Explanation

$\textbf{Step 1: Find the number of rectangles. }$

$\text{A rectangle on an } 8 \times 8 \text{ chess board can be of any dimension from } 1 \times 1, 1 \times 2, ...8 \times 8$

$\text{Total number of rectangle in an } n \times n \text{ chess board will be }$

$^{n+1}C_2 \times ^{n+1}C_2$

$\text{Here } n = 8$

$\therefore \text{Total number of rectangles = }$

$^{8+1}C_2 \times ^{8+1}C_2$

$=$

$^9C_2 \times ^9C_2$

$= 36 \times 36$

$= 1296$

$\textbf{Hence, the correct option is B.}$

The value of

$^nC_n$ is

0%
$n$
0%
$0$
0%
$1$
0%
$n!$

Explanation

Consider,

${ }^n{C}_{n}=\dfrac{n!}{(n-n)!n!}$

$=\dfrac{n!}{0!n!}$

$=1$ .................. since

$0!=1$

Hence,

${ }^n{C}_{n}=1$

$a, b, c\ \epsilon \left \{1, 2, .... 14\right \}$ . Let

$P(x) = ax^{2} + 2bx + c$ . What is the number of polynomials

$P(x)$ such that

$x + 1$ divides

$P(x)$ ?

$(a, b, c$ are distinct)

0%
$^{7}C_{2}$
0%
$^{7}C_{2}\cdot ^{7}C_{2}$
0%
$^{14}C_{3}$
0%
$2(^{7}C_{2} + ^{7}C_{2})$

Explanation

$P(x)=ax^2+2bx+c$

$,$

$P(-1)=0$

$\Rightarrow a-2b+c=0$

$\Rightarrow a+c=2b \rightarrow(1)$

$\therefore a,b,c$ are in

$A.P$

From

$(1)$ we know that

$a+c$ is even

$\Rightarrow a,b,c \in [1,2,........,14]$ Which contains

$7$ odd

$\xi$

$7$ even.

$\therefore$ Required

$={^7C_2}+{^7C_2}$

A man has 9 friends, 4 boys and 5 girls. In how many ways can be invite them, if there have to be exactly three girls in the invites?

0%
$320$
0%
$160$
0%
$80$
0%
$200$

Explanation

He can select three girls in

$^5C_3$ ways = 10 ways
He can select boys in

$[^4C_0+^4C_1 +^4C_2 +^4C_3 + ^4C_4]$ ways = 16 ways
Total number of ways

$= 10\times16$

$= 160$

The given table shows the possible food choices for lunch. How many different types of lunch can be made each including

$1$ type of soup,

$1$ type of sandwich and

$1$ type of salad?

Lunch Choices
Soup	Sandwich	Salad
Chicken	Cheese	Vegetable
Tomato	Paneer	Fruit

0%
$2$
0%
$3$
0%
$6$
0%
$8$

Explanation

Numbers of Soups

$= 2$
Number of Sandwiches

$= 2$
Number of Salads

$= 2$

$\therefore$ Required number of different types of lunch

$= 2\times 2\times 2 = 8$ .

The product of

$r$ consecutive integers is divisible by

$r!$ .

0%
True
0%
False

Explanation

The product of some r consecutive integers can be represented as

$\overbrace{(n+r)(n+r-1)\cdots(n+1)}^{r\mathrm{\ consective\ integers}}=\frac{(n+r)!}{n!}$

where n is the number one less than the smallest of the consecutive integers. Now, if it is true that primes in

$(n+r)!$ appear just as frequently or more as in n!r!, then you are saying that for some integer

$k$ (likely big) that

$(n+r)!=k⋅n!r!.$ So your product of n consecutive integers is

$\frac{(n+r)!}{n!}=\frac{k\cdot n!r!}{n!}=k\cdot r!$

and is therefore divisible by

$r!.$

CBSE Questions for Class 11 Engineering Maths Permutations And Combinations Quiz 1 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Permutations And Combinations Quiz 1 - MCQExams.com

0:0:1

Practice Class 11 Engineering Maths Quiz Questions and Answers

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