Class 11 Maths Chapter 7 Permutations and Combinations

Exercise 7.1

Question 1.
How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that
(i) repetition of the digits is allowed?
(ii) repetition of the digits is not allowed?
Solution.
There will be as many ways as there are ways of filling 3 vacant $\boxed { } \boxed { } \boxed { }$ places in succession by the five given digits.
(i) When repetition is allowed then each place can be filled in five different ways. Therefore, by the multiplication principle the required number of 3- digit numbers is 5 x 5 x 5 i.e., 125.
(ii) When repetition is not allowed then first place can be filled in 5 different ways, second place can be filled in 4 different ways & third place can be filled in 3 different ways. Therefore by the multiplication principle the required number of three digit numbers is 5 x 4 x 3 i.e, 60.

Question 2.
How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?
Solution.
There will be as many ways as there are ways of filling 3 vacant places $\boxed { } \boxed { } \boxed { }$ in succession by the 6 given digits. In this case we start filling in unit’s place, because the options for this place are 2, 4 & 6 only and this can be done in 3 ways. Ten’s and hundred’s place can be filled in 6 different ways. Therefore, by the multiplication principle, the required number of 3-digit even numbers is 6 x 6 x 3 i.e., 108.

Question 3.
How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?
Solution.
There will be as many ways as there are ways of filling 4 vacant places $\boxed { } \boxed { } \boxed { } \boxed { }$ in succession by the first 10 letters of the English alphabet, when repetition is not allowed then first place can be filled in 10 different ways, second place can be filled in 9 different ways, third place can be filled in 8 different ways and fourth place can be filled in 7 different ways. Therefore, by the multiplication principle the required number of 4 letter codes are 10 x 9 x 8 x 7 i.e., 5040.

Question 4.
How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?
Solution.
The 5 digit telephone numbers of the form $\boxed { 6 } \boxed { 7 }$ $\boxed { } \boxed { } \boxed { }$ can be constructed using the digits 0 to 9. When repetition is not allowed then at first & second place 6 & 7 are fixed respectively. Therefore, third, fourth and fifth place can be filled in 8, 7 and 6 ways respectively. So, by the multiplication principle the required numbers of 5-digit telephone numbers is 8 x 7 x 6 i.e., 336.

Question 5.
A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?
Solution.
When a coin is tossed there are two possible outcomes i.e. head or tail. When the coin is tossed three times then the total possible outcomes are 2 x 2 x 2 i.e., 8.

Question 6.
Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?
Solution.
There will be as many signals as there are ways of filling in 2 vacant places $\boxed { } \boxed { }$ in succession by the 5 flags of different colours. The upper vacant place can be filled in 5 different ways by anyone of the 5 flags ; following which the lower vacant place can be filled in 4 different ways by anyone of the remaining 4 different flags. Hence by the multiplication principle the required number of signals is 5 x 4 = 20.

Exercise 7.2

Question 1.
Evaluate
(i) 8!
(ii) 4!-3!
Solution.
(i) 8! = 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 = 40320
(ii) 4! – 3! = (1 x 2 x 3 x 4) – (1 x 2 x 3) = 24 – 6 = 18

Question 2.
Is 3! + 4! = 7! ?
Solution.
3! + 4! = (1 x 2 x 3) + (1 x 2 x 3 x 4) = 6 + 24 = 30 … (i)
7! = 1 x 2 x 3 x 4 x 5 x 6 x 7 = 5040 ………(ii)
From (i) & (ii), we get 3! + 4! ≠ 7!.

Question 3.
$\frac { 8! }{ 6!\times 2! }$
Solution.
$\frac { 8! }{ 6!\times 2! } =\frac { 8\times 7\times 6! }{ 6!\times 2! } =4\times 7=28$

Question 4.
$\frac { 1 }{ 6! } +\frac { 1 }{ 7! } =\frac { x }{ 8! }$ , findx.
Solution.
We have, $\frac { 1 }{ 6! } +\frac { 1 }{ 7! } =\frac { x }{ 8! }$
NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.2 1

Question 5.
Evaluate $\frac { n! }{ \left( n-r \right) ! }$ , when
(i) n = 6, r = 2
(ii) n = 9, r = 5
Solution.
(i) $\frac { 6! }{ \left( 6-2 \right) ! } =\frac { 6! }{ 4! } =\frac { 6\times 5\times 4! }{ 4! } =30$

(ii) $\frac { 9! }{ \left( 9-5 \right) ! } =\frac { 9! }{ 4! } =\frac { 9\times 8\times 7\times 6\times 5\times 4! }{ 4! } =15120$

Exercise 7.3

Question 1.
How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?
Solution.
Total digits are 9. We have to form 3 digit numbers without repetition.
∴ The required 3 digit numbers = 9P3
$=\frac { 9! }{ 6! } =9\times 8\times 7=504$

Question 2.
How many 4-digit numbers are there with no digit repeated?
Solution.
The 4-digit numbers are formed from digits 0 to 9. In four digit numbers 0 is not taken at thousand’s place, so thousand’s place can be filled in 9 different ways. After filling thousand’s place, 9 digits are left. The remaining three places can be filled in 9P3 ways.
So the required 4-digit numbers
= 9 x 9P3
= 9 x 504 = 4536.

Question 3.
How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?
Solution.
For 3-digit even numbers unit place can be filled by 2, 4, 6 i.e in 3 ways. Then the remaining two places can be filled in 5P2 ways.
∴ The required 3-digit even numbers
= 3 x 5P2
= 60

Question 4.
Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?
Solution.
The 4-digit numbers can be formed from digits 1 to 5 in 5P4ways.
∴ The required 4 digit numbers = 5P4 = 120 For 4-digit even numbers unit place can be filled by 2,4, i.e., in 2 ways. Then the remaining three places can be filled in 4P3 ways.
∴ The required 4-digit even numbers
= 2 x 4P3 = 2 x 24 = 48

Question 5.
From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person cannot hold more than one position?
Solution.
From a committee of 8 persons, we can choose a chairman and a vice chairman

Question 6.
Find n if n-1P3: nP4 = 1 : 9.
Solution.
NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.3 1

Question 7.
Find r if
(i) 5Pr = 26Pr-1
(ii) 5Pr = 6Pr-1
Solution.
vedantu class 11 maths Chapter 7 Permutations and Combinations Ex 7.3 2

Question 8.
How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?
Solution.
No. of letters in the word EQUATION = 8
∴ No. of words that can be formed
= 8P8 = 8!
=40320

Question 9.
How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if
(i) 4 letters are used at a time,
(ii) all letters are used at a time,
(iii) all letters are used but first letter is a vowel?
Solution.
No. of letters in the word MONDAY = 6
(i) When 4 letters are used at a time.
Then, the required number of words
= 6P4
$=\frac { 6! }{ 2! } =6\times 5\times 4\times 3=360$

(ii) When all letters are used at a time. Then the required number of words
= 6P6 = 6!
= 720

(iii) All letters are used but first letter is a vowel.
So the first letter can be either A or O.
So there are 2 ways to fill the first letter & remaining places can be filled in 5P5 ways.
∴ The required number of words
= 2 x 5P5
= 2 x 5! =240.

Question 10.
In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?
Solution.
There are 11 letters, of which I appears 4 times, S appears 4 times, P appears 2 times & M appears 1 time.
∴ The required number of arrangements
$=\frac { 11! }{ 4!4!2! } =\frac { 11\times 10\times 9\times 8\times 7\times 6\times 5\times 4! }{ 4\times 3\times 2\times 2\times 4! }$
= 10 x 10 x 9 x 7 x 5 = 34650 … (i)
When four I’s come together, we treat them as a single object. This single object with 7 remaining objects will account for 8 objects. These 8 objects in which there are 4S’s & 2P’s
can be rearranged in $\frac { 8! }{ 4!2! }$ ways i.e. in 840 ways … (ii)
Number of arrangements when four I’s do not come together = 34650 – 840 = 33810.

Question 11.
In how many ways can the letters of the word PERMUTATIONS be arranged if the
(i) words start with P and end with S,
(ii) vowels are all together,
(iii) there are always 4 letters between P and S?
Solution.
There are 12 letters of which T appears 2 times
(i) When words start with P and end with S, then there are 10 letters to be arranged of which T appears 2 times.
∴ The required words = $\frac { 10! }{ 2! }$
$=\frac { 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2! }{ 2! } =1814400$

(ii) When vowels are taken together i.e E U A I O we treat them as a single object. This single object with remaining 7 objects will account for 8 objects, in which there w are 2Ts, which can be rearranged in $\frac { 8! }{ 2! } =20160$ ways. Corresponding to each of these arrangements the 5 vowels E, U, A, I, O can be rearranged in 5! = 120 ways. Therefore, by multiplication principle, the required number of arrangements = 20160 x 120 = 2419200.

(iii) When there are always 4 letters between P & S
∴ P & S can be at
1st & 6th place
2nd & 7th place
3rd& 8th place
4th & 9th place
5th & 10th place
6th & 11th place
7th & 12th place.
So, P & S will be placed in 7 ways & can be arranged in 7 x 2! = 14
The remaining 10 letters with 2T’s, can be arranged in $\frac { 10! }{ 2! } =1814400$ ways.
∴ The required number of arrangements = 14 x 1814400= 25401600.

Exercise 7.4

Question 1.
lf nC8 = nC2, find nC2.
Solution.
We have, nC8 = nC2
NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.4 1

Question 2.
Determine n if
(i) 2nC3: nC3 =12 : 1
(ii) 2nC3: nC3= 11 : 1
Solution.
NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.4 2

Question 3.
How many chords can be drawn through 21 points on a circle?
Solution.
A chord is formed by joining two points on a circle.
∴ Required number of chords = 2nC2
$=\frac { 21! }{ 2!19! } =\frac { 21\times 20 }{ 2 } =210$

Question 4.
In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?
Solution.
3 boys can be selected from 5 boys in 5C3 ways & 3 girls can be selected from 4 girls in 4C3 ways.
∴ Required number of ways of team selection = 5C3 x 4C3 = $\frac { 5! }{ 2!3! } \times \frac { 4! }{ 3!1! }$
$=\frac { 5\times 4 }{ 2 } \times 4=40$

Question 5.
Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.
Solution.
No. of ways of selecting 3 red balls =6C3
No. of ways of selecting 3 white balls = 5C3
No. of ways of selecting 3 blue balls = 5C3
∴ Required no. of ways of selecting 9 balls
byjus class 11 maths Chapter 7 Permutations and Combinations Ex 7.4 4

Question 6.
Determine the number of 5 cards combinations out of a deck of 52 cards if there is exactly one ace in each combination.
Solution.
Total no. of cards = 52
No. of ace cards = 4
No. of non-ace cards = 48
∴ One ace card out of 4 can be selected in 4C1 ways.
Remaining 4 cards out of 48 cards can be selected in 48C4ways.
∴ Required no. of ways of selecting 5 cards
NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.4 5

Question 7.
In Kbw many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers?
Solution.
Total players = 17, No. of bowlers = 5,
No. of non-bowlers = 12
No. of ways of selecting 4 bowlers = 5C4
No. of ways of selecting 7 non-bowlers = 12C7
∴ Required no. of ways of selecting a cricket team
NCERT Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.4 6

Question 8.
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.
Solution.
No. of ways of selecting 2 black balls = 5C2
No. of ways of selecting 3 red balls = 6C3
∴ Required no. of ways of selecting 2 black & 3 red balls = 5C2 x 6C3
$=\frac { 5! }{ 2!3! } \times \frac { 6! }{ 3!3! } =\frac { 5\times 4 }{ 2 } \times \frac { 6\times 5\times 4 }{ 3\times 2 } =200$

Question 9.
In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?
Solution.
Total no. of courses = 9
No. of compulsory courses = 2
So, the student will choose 3 courses out of 7 courses [non compulsory courses].
∴ Required no. of ways a student can choose a programme = 7C3 = $\frac { 7! }{ 3!4! } =\frac { 7\times 6\times 5 }{ 6 } =35$

VERB AND FORMS OF VERB WITH HINDI MEANING_List of Verb in hindi

Verbs What is a verb? Verbs are the action words in a sentence that describe what the subject is doing. Along with nouns, verbs are the main part of a sentence or phrase, telling a story about what is taking place. Verbs are words that express action or state of being. There are three types of verbs: action verbs, linking verbs, and helping verbs . Action verbs are words that express action (give, eat, walk, etc.) or possession (have, own, etc.). Action verbs can be either transitive or intransitive. List of Verb in hindi Present Hindi Meaning Past Past Participle Buy खरीदना Bought Bought Build बांधना Built Built burn जलना Burnt Burnt Bend झुकना Bent Bent Bring लाना brought brought Become होना Became Become Come आना Came Come Catch पकड़ना Caught Caught Do करना Did Done Dream ख्वाब देखना Dreamt Dreamt Arise उठना / जागना Arose Arisen Be होना Was, were Been Bear सहन करना Bore Bore Beat मारना Beat Beat Bite काटना Bit Bitten Break तोडना Broke Broken Choose चुनना Chose Chosen Draw चि

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