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MATHS 3 Marks Question

Permutation and Combinations · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 22 questions.

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Question 13 Marks
Match each item given under the column C1 to its correct answer given under the column C2.
How many words (with or without dictionary meaning) can be made from the letters of the word MONDAY, assuming that no letter is repeated, if
C1
C2
(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
C1
C2
(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
Explanation:
  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
If nCr – 1 = 36, nCr = 84 and nCr + 1 = 126, then find rC2 .
[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 33 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 = 20C5 × 20C6 + 20C6 × 20C5 = 2 × 20C5 × 20C6
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Question 43 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 53 Marks
A convex polygon has 44 diagonals. Find the number of its sides.
[Hint: Polygon of n sides has (nC2 – n) number of diagonals.]
Answer
Let n be number of sides ploygon.
since, polygon of n sides has (nC2 - 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 63 Marks
Match each item given under the column C1 to its correct answer given under the column C2.
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 :
C1
C2
(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
C1
C2
(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
Explanation:
We have 3 books of mathematics, 4 of physics and 5 on english
  1. One book of each subject = 3C1 × 4C1 × 5C1 × = 3 × 4 × = 60
  2. Alleast one book of each subject = (22 - 1) × (24 - 1) × (25 - 1) = 7 × 15 × 31 = 3255
  3. Atlest one book of english = (25 - 1) × 27 = 31 × 128 = 3968. Hence the required matching is
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Question 73 Marks
There are 10 persons named P1 , P2 , P3 , ... P10. Out of 10 persons, 5 persons are to be arranged in a line such that in each arrangement P1 must occur whereas P4 and P5 do not occur. Find the number of such possible arrangements.
[Hint: Required number of arrangement = 7C4 × 5!]
Answer
Given that P1, P2, P3, P4, … P10 are 10 persons out of which 5 persons are to be arranged but P1 must occur and P4 and P5 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 × 5! = 35 × 120 = 4200
Hence, the required arrangement = 4200
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Question 83 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 93 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 = 16C11
  1. 9C14 If two particular players are included then more 9 players can be selected from remaining 14 players in 
  2. If two particular players are excluded then all 11 players can be selected from remaining 14 players in 14C11
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Question 103 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 = 18C2 – 5C2 + 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 = 18C2
Number of straight lines formed by joining 2 points out of 5 points = 5C2
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 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
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 = 5C2
And number of ways of selecting 3 red balls out of 6 red balls = 6C3
$\therefore$ Total number of ways of selecting 2 black and 3 red balls = 5C2 × 6C3
$=\frac{5.4}{2.1}\times\frac{6.5.4}{3.2.1}=10\times20=200$ Ways
Hence, the required ways of selecting the balls = 200.
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Question 133 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 143 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
  1. If they can be of any colour means we have to select 4 marbles out of 11

$\therefore$ Required number of ways = 11C4

  1. Two white marbles can be selected in 6C2 Two red marbles can be selected in 5C2 ways.

$\therefore$ Total number of ways = 6C2 × 5C2 = 15 × 10 = 150

  1. ways.4C5 ways. And 4 red marbles out of 5 can be selected in 4C6If they all must be of same colour, Four white marbles out of 6 can be selected in 

$\therefore$ Required number of ways = 6C4 + 5C4 = 15 + 5 = 20 

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Question 153 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 = 210 – 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 = 10C1 + 10C2 + 10C3… + 10C10
= 210 – 1 = 1024 - 1 = 1023
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Question 163 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 = 3C1 × 6C2 + 3C2 × 6C2 + 3C3 .]
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
= 3C1 × 6C2 + 3C2 × 6C1 + 3C3 × 6C0
=3 × 15 + 3 × 6 + 1 × 1 = 45 + 18 + 1 = 64
Hence, the required selection = 64.
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Question 173 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 183 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 193 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 8P4 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 × 1680 = 8400
Hence, the required telephone numbers = 8400
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Question 203 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
= 8-2C6-2 = 6C4 = 15
Case II: When A is not chosen.
$\therefore$ Number of ways = Number of ways of selecting 6 persons from remaining 7 person 7C6 = 7
$\therefore$ Total number of ways = 15 + 7 =22
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Question 213 Marks
Find the number of different words that can be formed from the letters of the word ‘TRIANGLE’ so that no vowels are together.
Answer
Total number of words in '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 = 6P3
$=\frac{6!}{(6-3)!}=\frac{6!}{3!}=120$ ways
so, the total arranment $=120\times120=14400$ ways
Here, the required arrangment = 14400 ways
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Question 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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