# NCERT Solutions for Class 11 Maths

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1.     (i) Given digits are 1, 2, 3, 4, 5 and repetition of digits are allowed.

(D3)   (D2)   (D1)

First digit (D1) can be arranged in 5 ways

Second digit (D2) can be arranged in 5 ways, because repetition is allowed.

Similarly, Third digit (D3) can be arranged in 5 ways

∴ Total number of three digit numbers 5 × 5 × 5 = 125

(ii) Given digits are 1, 2, 3, 4, 5 and repetition of digits are not allowed.

(D3)   (D2)   (D1)

First digit (D1) can be arranged in 5 ways

Second digit (D2) can be arranged in any of the remaining 4 digits because repetition is not allowed. So it can be done in 4 ways.

Similarly, Third digit (D3) can be arranged in 3 ways

∴ Total number of three digit numbers 5 × 4 × 3 = 60.

2.    Even numbers always end with a digit of 0, 2, 4, 6 or 8.

Given digits are 1, 2, 3, 4, 5, 6 and repetition of digits are allowed.

So unit place (D1) can be arranged by 2, 4 or 6.

(D3)   (D2)   (D1)

Second digit (D2) can be arranged in 6 ways because repetition is allowed. So it can be done in 6 ways.

Similarly, Third digit (D3) can be arranged in 6 ways

∴ Total number of three digit numbers 3 × 6 × 6 = 108.

3.    L1   L2   L3   L4

First letter (L1) can be arranged in 10 ways.

Second letter (L2) can be arranged in 9 ways, because repetition is not allowed.

Similarly, L3 and L4 can be arranged in 8 and 7 ways respectively . Total number of four letter code is 10 × 9 × 8 × 7 = 5040.

4.    6   7   D1   D2   D3

We need to create a five digit phone numbers which is always start with 67.

First two places have already filled with 67. Rest three places can be filled with digits from 0 to 9 also digits are not allowed to repeat.

Now, D3 can be arranged using digits 0 to 9 except 6 & 7.

Therefore D3 can be arranged in 8 ways.

D2 can be arranged in remaining 7 digits, so it can done in 7 ways.

D1 can be arranged in remaining 6 digits, so it can done in 6 ways.

∴ the total number of 5 – digit telephone numbers start with 67 is 9 × 7 × 6 = 336.

5.

When a coin is tossed, we get two outcomes either head or tail.

In each throw the number of ways of getting different outcome is 2.

Given that the coin is tossed three times,

So, using the multiplication principle, the number of possible outcome = 2 × 2 × 2 = 8.

6.    F1   F2

Given each signal requires the use of two flags namely F1 and F2.

We need to generate different signals, so repetition of flags are not allowed.

F1 can be arranged in 5 ways and F2 can be arranged in 4 ways because repetition of flags are not allowed.

By using multiplication principles , the total number of signals = 5 × 4 = 20.

7.    (i)8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40320

(ii) 4! – 3! = 4 × 3 × 2 × 1 – 3  × 2 × 1 = 24 – 6 = 18

8.    LHS = 3! + 4! = 3 × 2 × 1 + 4 × 3 × 2 × 1 = 6 + 24 = 30

RHS = 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040

∴ LHS ≠ RHS

9.    Given:

10.

11.   (i) Given: n = 6 and r = 2

12.   Given digits from 1 to 9 and repitition of digits are not allowed. We need to a three digit number.

9 different digits taken 3 at a time

13.   Thousand   Hundred   Tenth   Unit

Thousandth place can be arranged any of the digits from 1 to 9 because 0 cannot be included. So the number of ways to arrange thousandth place is 9.

The hundred, Tenths and Unit places can be arranged by any of the digits from 0 to 9 but digits cannot be repeated.

Thus the hundred, tenths and unit places can be arranged in 9 ways.

∴ there will be as many such 3 digit numbers as there are permutations 9 different digits taken 3 at a time.

∴ Number of such three digits

Thus by multiplication principle,

Total number of 4 digits numbers = 9 × 504 = 4536.

14.   Total number of digits = 6 and unit place can be filled with any one of the digits 2, 4, 6.

∴ Number of ways to arrange unit place

Now, the tens and hundreds places can be filled by remaining 5 digits.

∴ Number of ways to arrange the hundredth and tenth places

Therefore, total number of 3 – digit even numbers = 3 × 20 = 60.

15.   Thousand   Hundred   Tenth   Unit

Even numbers always end with a digit of 0, 2, 4, 6 or 8

Given digits are 1, 2, 3, 4, 5, Here even numbers end with 2 and 4.

The number of ways in which unit place can be arranged is 2

Since repetition of digits are not allowed, unit place is already arranged with one of the even numbers. So remaining places can be arranged with 4 digits.

Therefore, the number of ways to arrange remaining places is

Therefore, the required number of even numbers = 2 × 24 = 48.

16.   We need to form a committee of 8 people with a chairman and a vice – chairman. It is evident that no committee contains two or more chairman and two or more vice – chairman.

So repetition is not allowed.

Therefore, the number of ways to choose a chairman and a vice – chairman is the permutation of 8 things taken 2 at a time.

Required number of ways

17.

18.

r = 10, is not possible because r > n, therefore, r = 3

r = 9, is not possible because r > n, therefore, r = 4.

19.   Total number of letters in word ” EQUATION ” = 8

Number of letters to arranged = 8 (All distinct)

So it is a permutation of 8 distinct letters taken 8 at a time.

∴ Number of permutations

= 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40320.

20.   (i) We can arrange 6 – letter words from the letters on ” MONDAY ” without repetition.

It is the permuation of 6 distinct letters taken 6 at a time.

Therefore, required number of words

= 6 × 5 × 4 × 3 × 2 × 1 = 720.

(i) Total number of letters in word ” MONDAY ” = 6

Number of letters to arranged = 4 (All distinct)

So it is a permutation of 6 distinct letters taken 4 at a time.

∴ Number of permutations

= 6 × 5 × 4 × 3 = 360.

(iii) Total number of letters in word MONDAY = 6, four consonants M, N, D, Y and two vowels A and O.

There are two different ways to arrange the first letter. This can be done in ways. Since the letters cannot be repeated and the first letter is already arranged with one of the vowels, so the remaining letters can be arranged with 5 distinct letters. This can be done in 5P5 ways.

Therefore total number of words formed

= 5 × 4 × 3 × 2 × 1 × 2 × 1 = 240.

21.  In the word “MISSISSIPPI”, letter “I” appears 4 times, letter “S” appears 4 times, letter “P” appears 2 times, letter “M” appears once.

Therefore , the number of distinct permutations of the given word ” MISSISSIPPI ” is

Now, there are 4 I’s in the word “MISSISSIPPI”. When they occur together , treat them as a single letter. This single letter and the remaining 7 letters together we get total 8 letters.

In this 8 letters there are 4 “S” and 2 “P”, So it can be arranged in

Number of distinct permutations = 34650 – 840 = 33810.

22.   (i) In PERMUTATIONS number of letters = 12 and there are 2 T’s in it.

No. of ways letters of PERMUTATIONS be arranged so that words always start from P and end with S = = 1814400

(ii) There are 5 vowels in the word PERMUTATIONS and if they are together are considered them as 1, so no. of letters then is 8

No. of ways vowels are together

(iii) For 4 letters between P and S, P and S can be arranged in ways as shown:

P___S_____  __P____S___     _P_____S____   ___P___S___

____P____S__   _____P____S_   ______P_____S

As P can be filled in places 1, 2, 3, 4, 5, 6 and 7, consequently S can be filled in places 6,7,8,9,10,11 and 12 leaving 4 places between them. So P and S or S and P can be filled in 7 + 7 = 14 ways. Now the remaining 10 places can be filled in

Therefore No. of ways in which 4 letters between P and S = 1814400 × 14 = 2540100.

23.   Given: nC8 = nC2

⇒ nC3 = nC3

⇒ 8 = n – 2

⇒ n = 10

24.   (i) Given: nC3 : nC2 = 12 : 1

(ii) Given: nC3 : nC3 = 11 : 1

25.   There are 21 points on the circumference of a circle. Since one and only one chord can be drawn by joining 2 distinct points, therefore the required number of chords given by

26.  There are 5 boys and 4 girls and 3 out of 5 boys and 3 out of 4 girls have to be selected.

∴ Number of ways of selection = 5C3 × 4C3

27.   There are 6 red balls, 5 white balls and 5 blue balls. 3 balls of each colour have to be found.

∴ Number of ways of selection = 6C3 × 5C3 × 5C3

28.   There are 4 aces and 48 other cards in a deck of 52 cards. We have to select 1 ace out of 4 aces and 4 other cards out of 48 other cards.

∴ Number of ways of selection = 4C1 × 48C4

29.   There are 5 bowlers and 12 other players in a team of 17 players we have to select 4 bowels out of 5 bowlers and 7 other players out of 12 other players.

∴ Number of ways of selection = 5C4 × 12C7

30.   There are 5 black and 6 red balls. We have to select 2 black balls out of 5 black balls and 3 red balls out of 6 red balls.

∴ Number of ways of selection

31.  There are 9 courses and number of courses to be selected are 5 in which 2 specific courses are compulsory. 3 courses out of remaining 7 courses have to be selected.

∴ Number of ways of selection

32.  There are 8 letters in the word DAUGHTER. In this word 3 vowels and 5 consonants. 2 vowels and 3 consonants are to be selected.

∴ Number of ways of selection

Now, each word contains 5 letters which can be arranged among themselves 5! Ways.

Therefore, total number of words = 5! × 30 = 120 × 30 = 3600.

33.  There are 8 letters in the word EQUATION. In this word 5 vowels and 3 consonants.

Now, 5 vowels can be arranged in 5! Ways and 3 consonants can be arranged in 3! Ways.

Also the groups of vowels and consonants can be arranged in 2! Ways.

∴ Total number of permutations = 5! × 3! × 2! = 120 × 6 × 2 = 1440.

34.  (i) There are 9 boys and 4 girls. 3 girls and 4 boys have to be selected.

∴ Number of ways of selection

(ii) We have to select at least 3 girls. So the committee consists of 3 girls and 4 boys or 4 girls and 3 boys.

∴ Number of ways of selection

(iii) We have to select at most 3 girls. So the committee consists of no girls and 7 boys or 1 girl and 6 boys or 2 girls and 5 boys or 3 girls and 4 boys.

∴ Number of ways of selection

35.  In the word EXAMINATION, There are 11 letters, two A’s two I’s and two N’s and all other letters are different.

In Dictionary list, before E only A out of these letters can come.

After Fixing A as first letter, 10 letter are remaining out of we have two I’s And two N’s

∴ Number of ways of arrangement

36.   A number divisible by 10 have unit place digit 0. So digit 0 is fixed at unit place and the remaining 5 placed filled with remaining five digits in ways.

∴ Required numbers = 1 × nC3 = 1 × 5! = 120.

37.  2 vowels out of 5 vowels and 2 consonants out of 21 consonants have to be selected and these 4 letters in 4! ways.

∴ Required number of words

38.  Here, we have to select 8 questions at least 3 questions from each section. Therefore, we have required selections are 3 from part I and 5 from part II or 4 from part I and 4 from part II or 5 from part I and 3 from part II.

∴ Number of ways of selection

39.  Here, we have to select 5 cards containing 1 king and 4 other cards i.e., we have to select 1 king out of 4 kings and 4 other cards out of 48 other cards.

∴ Number of ways of selection = 4C1 × 48C4

40.  Given: Women occupy the even places.

M1 W1 M2 W2 M3 W3 M4 W4 M5

Therefore, we can arrange four women in 4! Ways and 5 men in 5! Ways.

∴ Number of ways of selection = 4! × 5! = 24 × 120 = 2880.

41.  According to the question, Number of ways of selection

42.   In the word ASSASSINATION, A appears 3 times, S appears 4 times, I appears in 2 times and N appears in 2 times. Now, 4 S’ taken together become a single letter and other remaining letters taken with this single letter.

Number of arrangements

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