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Maths 3 Marks Question

Permutation and Combinations · CBSE STD 11 Science (CBSE - English Medium). 22 questions.

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Question 13 Marks
Match each item given under the column $C_1$ to its correct answer given under the column $C_2.$
How many words $($with or without dictionary meaning$)$ can be made from the letters of the word $\text{MONDAY,}$ assuming that no letter is repeated$,$ if
 
$C_1$
 
$C_2$
$(a)$
$4$ letters are used at a time.
 $(i)$ $720$
$(b)$
All letters are used at a time.
 $(ii)$ $240$
$(c)$
All letters are used but the first is a vowel.
 $(iii)$ $360$
Answer
 
$C_1$
 
$C_2$
$(a)$
$4$ letters are used at a time.
 $(iii)$ $360$
$(b)$
All letters are used at a time.
 $(i)$ $720$
$(c)$
All letters are used but the first is a vowel.
 $(ii)$ $240$
  1. $4$ Letter are used at time $=\ ^6\text{P}_4=\frac{6!}{2!}=360$
  2. All letters are used at time $^6\text{P}_6=6!=720$
  3. All letters are used but first letter is vowel $=2\times5!=2\times120=240$
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Question 23 Marks
Match each item given under the column $C_1$ to its correct answer given under the column $C_2.$
There are $3$ books on Mathematics$, 4$ on Physics and $5$ on English. How many different collections can be made such that each collection consists of :
 
$C_1$
 
$C_2$
$(a)$
One book of each subject.
 $(i)$ $3968$
$(b)$
At least one book of each subject.
 $(ii)$ $60$
$(c)$
At least one book of English.
 $(iii)$ $3255$
Answer
 
 $C_1$
 
 $C_2$
$(a)$
One book of each subject.
 $(ii)$ $60$
$(b)$
At least one book of each subject.
 $(iii)$ $3255$
$(c)$
At least one book of English.
 $(i)$ $3968$
We have $3$ books of mathematics$, 4$ of physics and $5$ on english
  1. One book of each subject $=\ ^3C_1 \times\ ^4C_1 \times\ ^5C_1\ \times = 3 \times 4 \times = 60$
  2. Alleast one book of each subject $= (2^2 - 1) \times (2^4 - 1) \times (2^5 - 1) = 7 \times 15 \times 31 = 3255$
  3. Atlest one book of english $= (2^5 - 1) \times 2^7 = 31 \times 128 = 3968.$ Hence the required matching is
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Question 33 Marks
There are $10$ persons named $P_1 , P_2 , P_3 , ... P_{10.}$ Out of $10$ persons$, 5$ persons are to be arranged in a line such that in each arrangement $P_1$ must occur whereas $P_4$ and $P_5$ do not occur. Find the number of such possible arrangements.
$[$Hint: Required number of arrangement $ = ^7C_4 \times 5!]$
Answer
Given that $P_1, P_2, P_3, P_4, … P_{10}$ are $10$ persons out of which $5$ persons are to be arranged but $P_1$ must occur and $P_4$ and $P_5$ never occur.
$\therefore$ selection is to be done only for $10 – 3 = 7$ persons
$\therefore$ Number of selection $=\ ^7\text{C}_4\frac{7!}{4!(7-4)!}=\frac{7!}{4!3!}$
$=\frac{7.6.5.4!}{4!.3.2.1.}=35$
$5$ people can be aeeanment as $5!$
So$,$ the number of arrangement $= 35 \times 5! = 35 \times 120 = 4200$
Hence$,$ the required arrangement $= 4200$
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Question 43 Marks
In a certain city, all telephone numbers have six digits, the first two digits always being $41$ or $42$ or $46$ or $62$ or $64$. How many telephone numbers have all six digits distinct?
Answer
If first two digits is $41,$ then the remaining $4$ digits can be arranged in $^8P_4$ ways
$=\frac{8!}{(8-4)}=\frac{8\times7\times6\times5\times4!}{4!}=1680$
Similarly firth two digits can be $42$ or $46$ or $62$ or $64$
$\therefore$ Total number of telephone number have all digits distinct $= 5 \times 1680 = 8400$
Hence$,$ the required telephone numbers $= 8400$
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Question 53 Marks
In how many ways can a football team of $11$ players be selected from $16$ players? How many of them will.
  1. Include $2$ particular players?
  2. Exclude $2$ particular players?
Answer
Total number of players$ = 16$
We have to select a team of $11$ players
So$,$ number of ways $=\ ^{16}C_{11}$
$^9C_{14}^{ }$ If two particular players are included then more $9$ players can be selected from remaining $14$ players in
If two particular players are excluded then all $11$ players can be selected from remaining $14$ players in $^{14}C_{11}$
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Question 63 Marks
A candidate is required to answer 7 questions out of 12 questions, which are divided into two groups, each containing 6 questions. He is not permitted to attempt more than 5 questions from either group. Find the number of different ways of doing questions.
Answer
Since the candidate cannot attempt more than 5 questions from either group, he is able to attempt minimum two questions from either group.
The possible number of questions attempted from each group will be as given in the following table:
Group I
5
4
3
2
Group II
2
3
4
5
$\therefore$ Total nimber of possible ways $=2[^6\text{C}\times^6\text{C}_2+^6\text{C}_4\times^6\text{C}_3]$
$=2[6\times15\times+15\times20]=2\times390=780$
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Question 73 Marks
If 20 lines are drawn in a plane such that no two of them are parallel and no three are concurrent, in how many points will they intersect each other?
Answer
It is given that no two lines are parallel which means that all the lines are intersecting and no three lines are concurrent.
One point of intersection is created by two straight lines.
Number of points of intersection = Number of combinations of 20 straight lines taken two at a time
$=\ ^{20}\text{C}_2\frac{20!}{2!18!}=\frac{20\times19}{2\times1}=190$
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Question 83 Marks
Find the number of different words that can be formed from the letters of the word $\text{‘TRIANGLE’}$ so that no vowels are together.
Answer
Total number of words in $\text{'TRIANGLE'} = 8$
out of $5$ are consonants and $3$ are vowels
if vowels are not together, takan we have tha following arrangment
$V | C | V | C | V | C | V | C | V | C | V |$
Conjsonant can be arrngment in $5! = 120 $ways
Vowels occupy $6$ places
$\therefore 3$ vowels can be arranment in $6$ places $= ^6P_3$
$=\frac{6!}{(6-3)!}=\frac{6!}{3!}=120$ ways
so$,$ the total arranment $=120\times120=14400$ ways
Here$,$ the required arrangment $= 14400$ ways
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Question 93 Marks
Out of $18$ points in a plane, no three are in the same line except five points which are collinear. Find the number of lines that can be formed joining the point.
$[$Hint: Number of straight lines $= ^{18}C_2 – ^5C_2+ 1.]$
Answer
Total number of points $= 18$ Out of $18$ number$, 5$ are collinear and we get a straight line by joining any two points.
$\therefore$ Total number of straight line formed by joining $2$ points out of $18$ points $= ^{18}C_2$
Number of straight lines formed by joining $2$ points out of $5$ points $= ^5C_2$
But $5$ points are collinear and we get only one line when they are joined pair wise.
So$,$ the required number of straight lines are
$=\ ^{18}\text{C}_2-\ ^55_2+1=\frac{18.17}{2.1}-\frac{5.4}{2.1}+1=153-10+1=144$
Hence$,$ the total number of straight line $= 144$
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Question 103 Marks
A bag contains six white marbles and five red marbles. Find the number of ways in which four marbles can be drawn from the bag if:
  1. They can be of any colour.
  2. Two must be white and two red and.
  3. They must all be of the same colour.
Answer
Total number of marbles $= 6$ white $+- 5$ red $= 11$ marbles
If they can be of any colour means we have to select $4$ marbles out of $11$
$\therefore$ Required number of ways $=\ ^{11}C_4$
Two white marbles can be selected in $^6C_{2 }$Two red marbles can be selected in $^5C_2$ ways.
$\therefore$ Total number of ways $=\ ^6C_2 \times ^5C_2 = 15 \times 10 = 150$ ways.
$_4C^5$ ways. And $4$ red marbles out of $5$ can be selected in $_4C^6$ If they all must be of same colour$,$ Four white marbles out of $6$ can be selected in
$\therefore$ Required number of ways $=\ ^6C_4 + ^5C_4 = 15 + 5 = 20$
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Question 113 Marks
How many automobile license plates can be made if each plate contains two different letters followed by three different digits?
Answer
There are 26 English alphabets and 10 digits (0 to 9).
It is given that each plate contains two different letters followed by three different digits.
$\therefore$ Arrangment of 26 letters, taken 2 at a time $=\ ^{26}\text{P}_2=26\times25=650$ And arrangment of 10 digits, takan three at a time $=\ ^{10}\text{P}_3=10\times9\times8=720$
$\therefore$ Total number of licence plates $=650\times720=468000$
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Question 123 Marks
If $^nC_{r – 1 }= 36,^nC_r = 84$ and $^nC_{r + 1} = 126,$ then find $^rC_2 .$
$[$Hint: Form equation using $\frac{^\text{n}\text{C}_\text{r}}{^\text{n}\text{C}_{r+1}}$and $\frac{^\text{n}\text{C}_\text{r}}{^\text{n}\text{C}_{r-1}}$ to find the value of $r.]$
Answer
We know that $\frac{^\text{n}\text{C}_\text{r}}{^\text{n}\text{C}_\text{r}}=\frac{\text{n}-\text{r}+1}{\text{r}}$
$\therefore\ \frac{\text{n}-\text{r}+1}{\text{r}}=\frac{84}{36} ($given$)$
$\Rightarrow\ \frac{\text{n}-\text{r}+1}{\text{r}}=\frac{7}{3}$
$\Rightarrow\ 3_\text{n}-3\text{r}+3=7\text{r}$
$\Rightarrow\ 10\text{r}-3\text{n}=3\ ....(\text{i})$
$\frac{^\text{n}\text{C}_{\text{r}+1}}{^\text{n}\text{C}_\text{r}}=\frac{\text{n}-(\text{r}+1)+1}{\text{r}+1}=\frac{126}{84}$
$\therefore\ \frac{\text{n}-\text{r}}{\text{r}+1}=\frac{3}{2}$
$\Rightarrow\ 2\text{n}-2\text{r}=3\text{r}+3$
$\Rightarrow\ 2\text{n}-5\text{r}=3\ ....(\text{ii})$
Solving $(i)$ and $(ii),$ we get $n = 9$ and $r = 3.$
$\therefore\ ^\text{r}2_2=^3\text{C}_2=3$
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Question 133 Marks
Find the number of permutations of n distinct things taken r together, in which 3 particular things must occur together.
Answer
Total number of things. = n
We have to arrange r things out of n in which three particular things must occur together.
Therefore, combination of n things takan r time in which 3 things always occur $=\ ^{\text{n}-3}\text{C}_{\text{r}-3}$
three things takan together will be considered as 1 group.
Number of arrengement of three things $=3!$
Now Namber of arrangment of $\text{r}-3+1=(\text{r}-2)$ objects.
$\therefore$ number of arrangment of $(\text{r}-2)$ objects $=(\text{r}-2)!$
$\therefore$ Total no. of arranment $=\ ^{\text{n}-3}\text{C}_{\text{r}-3}\times(\text{r}-2)!\times3!$
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Question 143 Marks
A convex polygon has $44$ diagonals. Find the number of its sides.
$[$Hint: Polygon of n sides has $(^nC_2 – n)$ number of diagonals$.]$
Answer
Let n be number of sides ploygon.
since$,$ polygon of n sides has $(^nC_2 - n)$ number of diagonals
$\therefore\ ^\text{n}\text{C}_2-\text{n}=44=\frac{\text{n}!}{2!(\text{n}-2)!}-\text{n}=44$
$=\frac{\text{n}(\text{n-1})(\text{n-2})!}{2.(\text{n}-2)!}-\text{n}=44$
$\Rightarrow\ \frac{\text{n}(\text{n}-1)}{2}-\text{n}=44$
$=\frac{\text{n}^2-\text{n}-2\text{n}}{2}=44$
$\Rightarrow\ \text{n}^2-3\text{n}=88$
$\Rightarrow\ \text{n}^2-3\text{n}-88=0$
$=\text{n}^2-11\text{n}+8\text{n}-88=0$
$\Rightarrow\ \text{n}(\text{n}-11)=0$
$=(\text{n}-11)(\text{n}+8)=0$
$\therefore\ \text{n}=11$ and $\text{n}=-8[\because\text{n}\not=8]$ so $\text{n}=11$
Hence$,$ the required number of sides $= 11$
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Question 153 Marks
We wish to select $6$ persons from $8,$ but if the person $A$ is chosen$,$ then $B$ must be chosen. In how many ways can selections be made?
Answer
Total number of persons $= 8$
Case $I:$ When $A$ is chosen$, B$ must be chosen.
Number of whys $=$ Number of ways of selecting $4$ more persons from remaining $6$ persons
$= ^{8-2}C_6{-2} = \ ^6C_4 = 15$
Case $II:$ When A is not chosen.
$\therefore$ Number of ways $=$ Number of ways of selecting $6$ persons from remaining $7$ person $^7C_6 = 7$
$\therefore$ Total number of ways $= 15 + 7 =22$
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Question 163 Marks
A bag contains $5$ black and $6$ red balls. Determine the number of ways in which $2$ black and $3$ red balls can be selected from the lot.
Answer
Given that bag contains $5$ black and $6$ red balls.
Number of ways of selecting $2$ black ball out of $5$ black balls $= \ ^5C_2$
And number of ways of selecting $3$ red balls out of $6$ red balls $= \ ^6C_3$
$\therefore$ Total number of ways of selecting $2$ black and $3$ red balls $= \ ^5C_2 \times \ ^6C_3$
$=\frac{5.4}{2.1}\times\frac{6.5.4}{3.2.1}=10\times20=200$ Ways
Hence$,$ the required ways of selecting the balls $= 200.$
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Question 173 Marks
There are $10$ lamps in a hall. Each one of them can be switched on independently. Find the number of ways in which the hall can be illuminated.
$[$Hint: Required number $= 2^{10} – 1].$
Answer
There are $10$ lamps in a hall.
The hall can be illuminated if at least one lamp is switched.
$\therefore$ Total number of ways $= \ ^{10}C_{1 }+ ^{10}C_2 + ^{10}C_3… + ^{10}C_{10}$
$= 2^{10 }– 1 = 1024 - 1 = 1023$
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Question 183 Marks
18 mice were placed in two experimental groups and one control group, with all groups equally large. In how many ways can the mice be placed into three groups?
Answer
It is given that 18 mice were placed equally in two experimental groups and one control group i.e., three groups.
Each group is of 6 mice.
So, number of ways forming three group each of size $ '6' = \ ^{18}\text{C}_6\ ^{12}\text{C}_6\ ^6\text{C}_6​​​​​​​=\frac{18!}{6!\ 6!\ 6!}$
But in tish counting the order in wgich three particular groups are formed is also counted.
Clearly, when we from group the order must not be counted.
Now three particular groups can occur in 3! ways when order is counted.
So, actual number of ways $=\frac{18!}{6!\ 6!\ 6!}$
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Question 193 Marks
A box contains two white, three black and four red balls. In how many ways can three balls be drawn from the box$,$ if atleast one black ball is to be included in the draw.
$[$Hint: Required number of ways $=\ ^3C_1\times ^6C_2+ ^3C_2 \times ^6C_2 + ^3C_3 .]$
Answer
We have $2$ white$, 3$ black and $4$ red balls in a box. $3$ balls are to be drawn out of $9$ balls atleast one back ball is to be included So$,$ the possible selection is $(1$ black and $2$ other balls$)$ or $(2$ black and $1$ other ball$)$ or $(3$ black and no other ball$)$
So$,$ the number of possible selection is
$= ^3C_1 \times ^6C_2 + ^3C_2 \times ^6C_1 + ^3C_3 \times ^6C_0$
$=3 \times 15 + 3 \times 6 + 1 \times 1 = 45 + 18 + 1 = 64$
Hence$,$ the required selection $= 64.$
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Question 203 Marks
A group consists of $4$ girls and $7$ boys. In how many ways can a team of $5$ members be selected if the team has.
  1. No girls.
  2. At least one boy and one girl.
  3. At least three girls.
Answer
Number of girls $= 4$
Number of boys $= 7$
We have to select a team of $5$ members provided that
  1. Team having no. girls
$\therefore$ Required number of ways $=\ ^7\text{C}_5=\frac{7\times6}{2!}=21$
  1. Team having at least one boy and one girl
$\therefore$ Required number of ways
$=\ ^7\text{C}_1\times\ ^4\text{C}_4\ +\ ^7\text{C}_2\times\ ^4\text{C}_3+\ ^7\text{C}_3\times\ ^4\text{C}_2+\ ^7\text{C}_4\times\ ^4\text{C}_1$
$=7\times1+21\times4+35\times6+35\times4$
$=7+84+210+140$
$=441$
  1. Team having at least three girls
Required number of ways
 $=\ ^4\text{C}_3\times\ ^7\text{C}_2+\ ^4\text{C}_4\times\ ^7\text{C}_1$
$=4\times21+7$
$=84+7$
$=91$
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Question 213 Marks
A sports team of $11$ students is to be constituted$,$ choosing at least $5$ from Class $XI$ and atleast $5$ from Class $XII$. If there are $20$ students in each of these classes$,$ in how many ways can the team be constituted?
Answer
Total number of students in each class $= 20$
We have to select at least $5$ students from each class.
So we can select either $5$ students from class $XI$ and $6$ students from class $XII$ or $6$ students from class $XI$ and $5$ students from class $XII.$
$\therefore$ Total number of ways of selecting a team of $11$ players $= ^{20}C_5 \times ^{20}C_6 + ^{20}C_6 \times ^{20}C_5 = 2 \times ^{20}C_5 \times ^{20}C_6$
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Question 223 Marks
Find the number of integers greater than 7000 that can be formed with the digits 3, 5, 7, 8 and 9 where no digits are repeated.
[Hint: Besides 4 digit integers greater than 7000, five digit integers are always greater than 7000.]
Answer
Given that all the 5 digit numbers are greater than 7000.
So, the ways of forming 5-digit numbers = 5 × 4 × 3 × 2 × 1 = 120
Now all the four digit number greater than 7000 can be formed as follows.
Thousand place can be filled with 3 ways.
Hundred place can be filled with 4 ways.
Tenths place can be filled with 3 ways.
Units place can be filled with 2 ways.
So, the total number of 4- digit number = 3 × 4× 3 × 2 = 72
$\therefore$ Total number of integers = 120 + 72 = 192
Hence, the required number of integers = 192
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