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CBSE Questions for Class 11 Engineering Maths Permutations And Combinations Quiz 1 - MCQExams.com
CBSE
Class 11 Engineering Maths
Permutations And Combinations
Quiz 1
If $$\displaystyle { S }_{ { n } }=\sum _{ { r }=0 }^{ { n } } \frac { 1 }{ ^{ n }{ { C }_{ r } } } $$ and $$\displaystyle { t }_{ { n } }=\sum _{ { r }=0 }^{ { n } } \frac { { r } }{ ^{ n }{ { C }_{ r } } } $$, then $$\displaystyle \frac { { t }_{ { n } } }{ { s }_{ { n } } } =$$
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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
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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.
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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: $$\displaystyle \sum_{r=0}^{n}(r+1)^{n}C_{r}x^{r}=(1+x)^{n}+nx(1+x)^{n-1}$$
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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
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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
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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
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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
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0%
$$ ^{20}P_8 $$
0%
$$ ^{20}C_8 $$
0%
$$ 512 $$
0%
$$ 256 $$
Explanation
$$8$$ out of the $$20$$ words have an alternative solution.
So $$12$$ 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 $$x = $$
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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
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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
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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
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$$20$$
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
4
4
candidates for a Natural science scholarship,
2
2
for a Classical and
6
6
for a Mathematical scholarship,then find the n
o. of ways one of these scholarship can be awarded is,
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$$6$$
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.
$$n!=n(n-1)(n-2).....3.2.1$$
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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?
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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.
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0%
True
0%
False
0%
Either
0%
Neither
Explanation
Factorial : product of an integer with integer less than it.
Factorial can be interpolated using gamma function and gamma function and
gamma function is not defined for negative integer.
Factorial is not defined for negative integer
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:
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0%
19670
0%
39758
0%
69760
0%
99748
Explanation
(C) The total number of words that can be formed with five letters out of the ten given letters
=
10
5
=
100000
=
10
5
=
100000
Value of 0! is always 1.
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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.
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0%
True
0%
False
0%
Either
0%
Neither
Explanation
In combination
Each combination can be considered as a set of selection an order
Each selection has a defined order
They can be considered as a permutation
Each cpmbination corresponds to many permutations
Hence the above statement is true.
The value of $$\displaystyle ^{ 6 }C_{ 4 }$$ is
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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:
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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
Report Question
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$$9$$
0%
$$10$$
0%
$$11$$
0%
$$12$$
Explanation
The possible two-digit number are
88, 85, 82, 81, 87, 86, 58, 28, 18, 78, 68
These are 11 in number
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.
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0%
156
0%
144
0%
132
0%
72
0%
66
Explanation
There being 12 friends in the group, each friend must have purchased (12 - 1) i.e. 11 cards for sending greeting to rest of his 11 friends. Thus total number of cards purchased by all the friends together is 12 x 11 i.e. 132.
'
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?
Report Question
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 $$7,0,9,8,6,3$$
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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
Report Question
0%
One
0%
Three
0%
Five
0%
Six
Explanation
There are 3 such 9s that are not immediately preceded by 5 and immediately followed by 2 or 3 in the given series. They are marked in bold.
1 9 2
6 5 9 3 8
3 9 3
2 5 9
2 9 3
4 8 2 6 9 8
Answer is Option B
How many numbers of four digits can be formed from the digits 0, 1, 2, 3, and 4?
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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 = $$\displaystyle 4\times 4\times 3\times 2=96$$
How many numbers amongst the numbers 9 to 54 are there which are exactly divisible by 9 but not by 3?
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0%
8
0%
6
0%
5
0%
Nil
Explanation
Any number divisible by 9 is also divisible by 3
Arrange the given words in the sequence in which they occur in the dictionary and then choose the correct sequence.
PagePaganPalisadePageant
Palate
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0%
$$1,4,2,3,5$$
0%
$$2,4,1,3,5$$
0%
$$2,1,4,5,3$$
0%
$$1,4,2,5,3$$
Explanation
Words 1,2 and 4 have first 3 letter as "pag" and words "3 and 5 have first 3 letters as "pal". So words starting with "pal" will come later than words starting with "pag" as in Alphabet set the letter "g" comes after letter "a".
The fourth letter of word 5 is "a" and of word 3 is "i" hence word 3 will be the last in order.
So choice A and B are incorrect and the answer will be either C or D.
Simply arranging the 4th letter of words 1,2 and 4 will give the correct sequence choice as C.
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?
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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
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0%
26
0%
28
0%
30
0%
32
Explanation
Total number of students=100
Number of students fail in English=(50-12)=38
Number of students fail in Maths=(30-12)=18
Number of students fail in both=12
Therefore total failing students=(38+18+12)=68
Pass in both the subjects=(100-68)
=32 students
32 students have passed in both subjects
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?
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0%
20
0%
45
0%
55
0%
90
Explanation
Total number of handshakes
=
10
×
9
2
=
45
\dAXisplaystyle=10×92=45$$
A group of 1200 persons consisting of captains and soldiers is travelling in a train. For every 15 soldiers there is one captain. The number of captains in the group is:
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0%
85
0%
80
0%
75
0%
70
Explanation
Out of 16 men, there is a captain.
Number of captains in 1200 men = $$\displaystyle \\ 1200\div 16=75$$.
A car driver knows four different routes from Delhi to Amritsar. From Amritsar to Pathankot, he knows three different routes and from Pathankot to Jammu he knows two different routes. How many routes does he know
from Delhi to Jammu?
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0%
4
0%
8
0%
12
0%
24
0%
36
Explanation
The car driver can reach Amritsar in 4 ways.From each of these four ways, he can reach Pathankot in 3 different ways and so he can reach from Delhi to Pathankot in $$\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) $$ i.e. in 24 ways.
In a chess tournament each of six players will play every other player exactly once. How many matches will be played during the tournament?
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0%
36
0%
30
0%
15
0%
12
Explanation
First player can play 5 matches with other five players. Second player can play 4 matches with other four players and proceeding this way, the fifth player will
play only one match with sixth player.
$$\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 } $$
Amy and Adam are making boxes of truffles to give out as wedding favors. They have an unlimited supply of 5 different types of truffles. If each box holds 2 truffles of different types, how many different boxes can they make?
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0%
$$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$$.
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0%
$$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$$?
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0%
$$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 }$$?
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0%
$$\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}} $$
A bag contains Rs. $$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$$-paise coins?
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0%
$$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$$.
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0%
$$-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$$.
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0%
$$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$$ women
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0%
$$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$$
If $$^{20}C_{r+2} = ^{20}C_{2r-3}$$, then find $$^{12}C_r$$
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0%
$$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$$
The number of rectangles that you can find on a chess board is :
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0%
$$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
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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)
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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?
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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
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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 $$r!$$.
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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 $$r!.$$
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Practice Class 11 Engineering Maths Quiz Questions and Answers
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