CBSE Class 11 Mathematics Permutations And Combinations Worksheet Set 03

Question. n – digit numbers are formed using only three digits 2, 5 and 7. The smallest value of n for which 900 such distinct numbers can be formed, is 
(a) 6
(b) 8
(c) 9
(d) 7
Answer : D

Question. 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 :
(a) 44th
(b) 45th
(c) 46th
(d) 47th
Answer : C

Question. If n = ^mC₂, then the value of ⁿC₂ is given by
(a) 3(^{m + 1}C₄)
(b) ^{m – 1}C₄
(c) ^{m + 1}C₄
(d) 2(^{m + 2}C₄)
Answer : A

Question. The number of ways in which an examiner can assign 30 marks to 8 questions, giving not less than 2 marks to any question, is :
(a) ³⁰C₇
(b) ²¹C₈
(c) ²¹C₇
(d) ³⁰C₈
Answer : C

Question. A commit tee of 4 persons is to be formed from 2 ladies, 2 old men and 4 young men such that it includes at least 1 lady, at least 1 old man and at most 2 young men.
Then the total number of ways in which this committee can be formed is :
(a) 40
(b) 41
(c) 16
(d) 32
Answer : B

Question. Statement 1: If A and B be two sets having p and q elements respectively, where q > p. Then the total number of functions from set A to set B is q_p. 
Statement 2: The total number of selections of p different objects out of q ob ects is ^qC_p.
(a) Statement 1 is true, Statement 2 is false.
(b) Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation of Statement 1.
(c) Statement 1 is false, Statement 2 is true
(d) Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation of Statement 1.
Answer : D

Question. From 6 different novels and 3 different dictionaries,4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle.
Then the number of such arrangement is:
(a) at least 500 but less than 750
(b) at least 750 but less than 1000
(c) at least 1000
(d) less than 500
Answer : C

Question. How many different words can be formed by umbling the letters in the word MISSISSIPPI in which no two S are ad acent?
(a) 8. ⁶C₄. ⁷C₄
(b) 6.7. ⁸C₄
(c) 6. 8. ⁷C₄.
(d) 7. ⁶C₄. ⁸C₄
Answer : D

Question. Five digit number divisible by 3 is formed using 0, 1, 2, 3, 4, 6 and 7 without repetition. Total number of such numbers are
(a) 312
(b) 3125
(c) 120
(d) 216
Answer : D

Question. Two women and some men participated in a chess tournament in which every participant played two games
with each of the other participants. If the number of games that the men played between themselves exceeds the number of games that the men played with the women by 66, then the number of men who participated in the tournament lies in the interval:
(a) [8, 9]
(b) [10, 12)
(c) (11, 13]
(d) (14, 17)
Answer : B

Question. The number of integers greater than 6,000 that can be formed, using the digits 3, 5, 6, 7 and 8, without repetition, is :
(a) 120 (b) 72
(c) 216 (d) 192 20. The number of ways of selecting 15 teams from 15 men and 15 women, such that each team consists of a man and a woman, is:
(a) 1120 (b) 1880
(c) 1960 (d) 1240
Answer : D

Question. If all the words (with or without meaning) having five letters, formed using the letters of the word SMALL and arranged as in a dictionary; then the position of the word SMALL is :
(a) 52nd
(b) 58th
(c) 46th
(d) 59th
Answer : B

Question. If seven women and seven men are to be seated around a circular table such that there is a man on either side of every woman, then the number of seating arrangements is
(a) 6! 7!
(b) (6!)2
(c) (7!)2
(d) 7!
Answer : A

Question. 8-digit numbers are formed using the digits 1, 1, 2, 2, 2, 3, 4, 4. The number of such numbers in which the odd digits do no occupy odd places, is:
(a) 160
(b) 120
(c) 60
(d) 48
Answer : B

Question. The sum of the digits in the unit’s place of all the 4-digit numbers formed by using the numbers 3, 4, 5 and 6, without repetition, is: 
(a) 432
(b) 108
(c) 36
(d) 18
Answer : B

Question. If the four letter words (need not be meaningful) are to be formed using the letters from the word "MEDITERRANEAN" such that the first letter is R and the fourth letter is E, then the total number of all such words is :
(a) 110
(b) 59
(c) 11!/(2!)³
(d) 56
Answer : B

Question. Assuming the balls to be identical except for difference in colours, the number of ways in which one or more balls can be selected from 10 white, 9 green and 7 black balls is:
(a) 880
(b) 629
(c) 630
(d) 879
Answer : D

Question. The number of ways in which 5 boys and 3 girls can be seated on a round table if a particular boy B1 and a particular girl G1 never sit ad acent to each other, is : 
(a) 5 × 6!
(b) 6 × 6!
(c) 7!
(d) 5 × 7!
Answer : A

Question. The number of points, having both co-ordinates as integers, that lie in the interior of the triangle with vertices (0, 0), (0, 41) and (41, 0) is :
(a) 820
(b) 780
(c) 901
(d) 861
Answer : B

Question. 5 - digit numbers are to be formed using 2, 3, 5, 7, 9 without repeating the digits. If p be the number of such numbers that exceed 20000 and q be the number of those that lie between 30000 and 90000, then p : q is :
(a) 6 : 5
(b) 3 : 2
(c) 4 : 3
(d) 5 : 3
Answer : D

Question. An eight digit number divisible by 9 is to be formed using digits from 0 to 9 without repeating the digits. The number of ways in which this can be done is:
(a) 72 (7!)
(b) 18 (7!)
(c) 40 (7!)
(d) 36 (7!)
Answer : D

Question. If the letters of the word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word SACHIN appears at serial number
(a) 601
(b) 600
(c) 603
(d) 602
Answer : A

Q1. How many 3-digit number can be formed without using digits 1, 2, 3, 9, 5 if repetition of digits is not allowed ?
(Ans. 48)

Q2. Find n if ⁹P₅ + 5.⁹P₄ = ¹⁰P_n

Q3. How many numbers can be formed from the digits 2, 4, 6, 9 if repetition of digits is not allowed?
(An. 64)

Q4. Evaluate n, if P(15, n – 1) : P (16, n – 2) = 3:4 (Ans 14)

Q5. Prove that 2n!/n! = [1.3.5 ………. (2n-1)] 2ⁿ

Q6. 8 Children are to be seated on a bench
(i) In how many ways can the children be seated? (8! = 40320)
(ii) How many arrangements are possible if the youngest child sits at the left hand end of the bench? (50:40)

Q7. From 30 teachers in school, 1 Principal & 1 Vice Principal are to be appointed. In how many ways can this be done?

Q8. Find total number of ways of answering 5 objectives type questions each questions having 4 choices.
(Ans 45)

Q9. How many numbers are there between 100 and 1000 such that 5 is in the unit place?
(Ans 90)

Q10. A gentleman has 6 friends to invite. In how many ways can he send invitation cards to them if he has 3 servants to carry the cards?
(Ans. 729)

Q11. 3 married couples are to be seated in a row having 6 seats in a cinema hall. If spouses are to be seated next to each other, in how many ways can they be seated? Find also number of ways of their seating if all the ladies together.

Q12. How many different words can be formed with the letters of the word HARYANA ? How many of these
(i) Have H & N together?
(Ans. 120) (ii) Begin with H and end with N? (Ans. 20)
(iii) Have three vowels together?
(Ans. 120)

Q13. How many numbers are there between 100 and 1000 such that every digit is either 2 or 9 ? (Ans 8)

Q14. Prove that ⁿc_r + ⁿc_r-1 = ⁿ⁺¹c_r

Q15. If ⁸C_r – ⁷C₃ = ⁷C₂, find the value of r (3 or 5)

Q16. A committee of 5 is to be selected from among 6 boys and 5 girls. Determine the no. of ways of selecting the committee if it is to consist of at least one boy and one girl
(Ans. 455)

Q17. How many different words, each containing 2 vowels and 3 consonants can be formed with 5 vowels and 17 consonants.
(Ans. 816000)

Q18. In a meeting after everyone had shaken hands with everyone else, it was found that 66 handshakes were exchanged. How many members were present at the meeting.
(Ans. 12)

Q19. If a polygon has 44 diagonals, then what is the number of its sides
(Ans. 11)

Q20. If ¹⁶C_r = ¹⁶C_r+2, find rc4.
(Ans. 35)

Q21. If ⁿP_r = ⁿP_r+1 and ncr = ncr-1, find the values of n and r. (Ans n = 3, r = 2)

Q22. If ⁿ⁺²C₈ : ^n-2C₄ = 57:16, find n (n = 19).

Q23. If α = ^mc₂ then fixed ^αC₂

Q24. Find the number of ways in which 5 boys and 5 girls be seated in a row such that:-
(i) No two girls may sit together (5! X 6!
(ii) All the girls sit together and all the boys sit together (2 x 5! 5!)
(iii) All the girls are never together (10! – 5! X 6!)

Q.1 If ⁿC₇ = ⁿC₄, find the value of n.

Q.2 If 1/6! + 1/7! = x/8!, find x

Q.3 Out of 3 books on economics, 4 books on political science and 5 books on Geography, how many collections can be made if each collection consist of exactly one book on each subject?

Q.4 How many numbers greater than 1000000 can be formed by using the digits 1, 2, 0, 2, 4, 2, 4,?

Q.5 Find the value of n such that ⁿP₅ = 42ⁿP₃ , n > 4.

Q.6 How many 3-digit numbers can be formed from the digit 1, 2, 3, 4 and 5 assuming that (i) repetition of the digit is allowed? (ii) repetition of the digits is not allowed?

Q.7 In an examination, a question paper consists of 12 questions divided into parts, i.e. Part I and Part II, containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions?

Q.8 How many 4-digit numbers are there with no digit repeated?

Q.9 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 at least one boy and one girl?

Q.10 How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 5, 6, if no digit is repeated?

Q.11 What is number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of these (i) four cards are of the same suits, (ii) are face cards.

Q.12 How many words with or without meaning each of 3 vowels and 2 consonants can be formed from the letters of the word INVOLUTE?

Q.13 Eighteen guests have to be seated, half on each side of a long table. Four particular guests desire to sit on one particular side and three others on the other on side. Determine the number of ways in which the sitting arrangement can be made.

Q.14 In how many of the distinct permutations of the letters in MISSISSIPPI do the four 1’s not come together?

Q.15 Find the number of different 8-letters arrangements that can be made from the letters of the word DAUGHTER so that (i) all vowels occur together, (ii) all vowels do not occur together.

Q.16 If ⁿC₂ - ⁿC₁ = 35, then find the value of n.

Q.17 In a class there are 27 boys and 15 girls. The teacher wants to select a boys and a girl for the monitor ship of the class. In how many ways can the teacher make this selection?

Q.18 A person has got 15 acquaintances of whom 10 are relatives. In how many ways he may invite 9 guests so that 7 of them would be relatives?

Q.19 A committee of 3 persons is to be constituted from a group of 2 men and 3 women. In how many ways can this be done? How many of these committees would consist of 1 man and 2 women?

Q.20 A box contains 7 red 6 white and 4 blue balls. How many selection of three balls can be made so that all three are red balls?

BASICS:-

Permutations: Number of ways of arrangement of objects.

Combinations: Number of ways of selection of objects.

\[ ^nP_r = \frac{n!}{(n - r)!} \]

where:
\( n \rightarrow \) Number of items/objects available.
\( r \rightarrow \) Number of items/objects to be arranged.

\[ ^nC_r = \frac{n!}{r!(n - r)!} \]

where:
\( n \rightarrow \) Number of items/objects available.
\( r \rightarrow \) Number of items/objects to be selected.

Relation between \( ^nP_r \) and \( ^nC_r \):-
\( ^nP_r = {^nC_r} \times r! \)

Note: Mainly two operations:-
Addition (+): Used for "or" options and cases.
Multiplication (\(\times\)): Used for "and"/compulsion where the selection or arrangement is not yet completed.

Shortcuts of \( ^nC_r \):-

• \( ^nC_0 = 1 \qquad \text{E.g., } ^7C_0 = 1 \)
• \( ^nC_1 = n \qquad \text{E.g., } ^7C_1 = 7 \)
• \( ^nC_2 = \frac{n(n-1)}{2} \qquad \text{E.g., } ^7C_2 = \frac{7 \times 6}{2} = 21 \)
• \( ^nC_3 = \frac{n(n-1)(n-2)}{6} \qquad \text{E.g., } ^7C_3 = \frac{7 \times 6 \times 5}{6} = 35 \)
• \( ^nC_n = 1 \qquad \text{E.g., } ^7C_7 = 1 \)
• \( ^nC_r = {^nC_{n-r}} \qquad \text{E.g., } ^{10}C_8 = {^{10}C_2} \); \( ^{20}C_{19} = {^{20}C_1} \)
• If \( ^nC_x = {^nC_y} \), then \( x = y \) or \( x + y = n \).

Question. How many 3 digit even numbers can be made using the digits 1,2,3,4,6,7. If no digit is repeated?
Answer: Digits available: 1, 2, 3, 4, 6, 7
Required: 3-digit even numbers.
To form an even number, the units place can only be filled by the even digits \( (2, 4, 6) \).

1. The units place can be filled in 3 ways.
2. Since repetition is not allowed, the hundreds place can be filled by any of the remaining 5 digits (5 ways).
3. The tens place can then be filled in 4 ways.

\( \therefore \) The required number of 3-digit even numbers that can be formed \( = 5 \times 4 \times 3 = 60 \) ans.

Question. How many 6-digit numbers can be formed from the digits 0,1,3,5,7,9 . Which are divisible by 10 when:-
(i) Repeat of digits not allowed
(ii) Repeat of digits allowed
Answer: Digits available: 0, 1, 3, 5, 7, 9.
Required: 6-digit numbers divisible by 10.

1. When repetition of digits is not allowed:
- For numbers to be divisible by 10, the units place must be filled with 0 (1 way).
- The remaining 5 places can be filled by the 5 remaining non-zero digits in \( 5 \times 4 \times 3 \times 2 \times 1 \) ways respectively.
\( \therefore \) The required 6-digit numbers divisible by 10 when repetition is not allowed \( = 5 \times 4 \times 3 \times 2 \times 1 \times 1 = 120 \) ans.

2. When repetition of digits is allowed:
- The units place must be 0 (1 way).
- The first place (hundred-thousands) cannot be 0, so it can be filled in 5 ways.
- Each of the other 4 intermediate places can be filled by any of the 6 available digits (6 ways each).
\( \therefore \) The required numbers \( = 5 \times 6 \times 6 \times 6 \times 6 \times 1 = 6480 \) ans.

Question. How many 4-digit numbers divisible by 4 can be made with the digits 1,2,3,4,5. If repetitions of digits not allowed?
Answer: Digits available: 1, 2, 3, 4, 5.
Required: 4-digit numbers divisible by 4.

1. For a number to be divisible by 4, the number formed by its last two digits must be divisible by 4.
2. Out of the given digits, the possible divisible last two digits are: 12, 32, 24, and 52.
3. Thus, there are 4 cases:
- (i) Numbers ending with 12:
The remaining 2 places can be filled by the remaining 3 digits in \( 3 \times 2 = 6 \) ways.
- (ii) Numbers ending with 24:
Filled in \( 3 \times 2 = 6 \) ways.
- (iii) Numbers ending with 32:
Filled in \( 3 \times 2 = 6 \) ways.
- (iv) Numbers ending with 52:
Filled in \( 3 \times 2 = 6 \) ways.

\( \therefore \) The required numbers divisible by 4 \( = 6 + 6 + 6 + 6 = 24 \) ans.

Question. How many numbers between 100 and 1000 which have exactly one of their digit is 7?
Answer: Required: 3-digit numbers having exactly one digit as 7.
Digits available: 0 to 9.
Since it is not specified, repetition of digits is allowed.
There are 3 cases:

1. Let 7 be in the units place:
- Units place is 7 (1 way).
- Hundreds place can be filled in 8 ways (excluding 0 and 7).
- Tens place can be filled in 9 ways (excluding 7).
Number of ways \( = 8 \times 9 \times 1 = 72 \).

2. Let 7 be in the tens place:
- Tens place is 7 (1 way).
- Hundreds place can be filled in 8 ways (excluding 0 and 7).
- Units place can be filled in 9 ways (excluding 7).
Number of ways \( = 8 \times 1 \times 9 = 72 \).

3. Let 7 be in the hundreds place:
- Hundreds place is 7 (1 way).
- Tens place can be filled in 9 ways (excluding 7).
- Units place can be filled in 9 ways (excluding 7).
Number of ways \( = 1 \times 9 \times 9 = 81 \).

\( \therefore \) The required numbers \( = 72 + 72 + 81 = 225 \) ans.

Question. How many numbers are there between 100 and 1000 such that atleast one of their digit is 7 ?
Answer: The required numbers must be 3-digit numbers.
Digits available: 0 to 9.
Repetition of digits is allowed.

Required 3-digit numbers having at least one digit as 7 \( = \) (Total 3-digit numbers) \(-\) (3-digit numbers in which 7 does not appear at all).

(i) Total 3-digit numbers (from 100 to 999):
\( 9 \times 10 \times 10 = 900 \).

(ii) 3-digit numbers in which 7 does not appear at all:
- Hundreds place can be filled in 8 ways (excluding 0 and 7).
- Tens place can be filled in 9 ways (excluding 7).
- Units place can be filled in 9 ways (excluding 7).
Number of ways \( = 8 \times 9 \times 9 = 648 \).
Excluding the number 100 itself (which has no 7) from the range bounds:
Total 3-digit numbers in the range \( = 900 - 1 = 899 \).
3-digit numbers without 7 in the range \( = 648 - 1 = 647 \).

\( \therefore \) The required numbers \( = 899 - 647 = 252 \) ans.

Question. How many 3 digit even numbers can be found such that if 5 is one of the digit then 7 must be the next digit?
Answer: Required: 3-digit even numbers.
Digits available: 0 to 9.
There are two cases:

Case 1: When the digit 5 is present:
If 5 is present, it must be followed immediately by 7. To keep the number even, 5 cannot be in the tens place (since 7 would then be in the units place, making it odd). Likewise, 5 cannot be in the units place. Thus, 5 can only occupy the hundreds place, forcing 7 into the tens place:
- Hundreds place is 5 (1 way).
- Tens place is 7 (1 way).
- Units place can be filled by any even digit \( \{0, 2, 4, 6, 8\} \) (5 ways).
Number of ways \( = 1 \times 1 \times 5 = 5 \).

Case 2: When the digit 5 is not present:
- Hundreds place can be filled in 8 ways (excluding 0 and 5).
- Tens place can be filled in 9 ways (excluding 5).
- Units place can be filled by any even digit \( \{0, 2, 4, 6, 8\} \) (5 ways).
Number of ways \( = 8 \times 9 \times 5 = 360 \).

\( \therefore \) The required 3-digit numbers \( = 5 + 360 = 365 \) ans.

Question. Find the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time.
Answer: Digits available: 2, 3, 4, 5.
Required: 4-digit numbers formed using all available digits without repetition.

(i) Total 4-digit numbers that can be formed \( = 4! = 24 \).

(ii) In the units place, each digit (2, 3, 4, 5) occurs \( 3! = 6 \) times.
\( \therefore \) Sum of digits in the units place:
\( = 6 \times (2 + 3 + 4 + 5) = 6 \times 14 = 84 \).

(iii) Similarly, the sum of all digits in the tens, hundreds, and thousands places will each be 84.

(iv) The total sum of all 24 numbers is given by:
\( = 84 \times (10^3 + 10^2 + 10^1 + 10^0) \)
\( = 84 \times (1000 + 100 + 10 + 1) \)
\( = 84 \times 1111 = 93324 \) ans.

Question. How many numbers greater than 1000000 can be formed using the digits 1,2,0,2,4,2,4?
Answer: Required: 7-digit numbers.
Digits available: 1, 2, 0, 2, 4, 2, 4 (7 digits: 2 repeats 3 times, 4 repeats 2 times, 1 and 0 occur once).

The required numbers \( = \) (Total number of 7-digit numbers) \(-\) (Numbers starting with 0).

(i) Total possible 7-digit numbers:
\( = \frac{7!}{3! \times 2!} = \frac{5040}{6 \times 2} = 420 \).

(ii) Number of arrangements starting with '0':
If '0' is fixed in the first place, the remaining 6 digits can be arranged in:
\( = \frac{6!}{3! \times 2!} = \frac{720}{6 \times 2} = 60 \text{ ways} \).

\( \therefore \) The required numbers \( = 420 - 60 = 360 \) ans.

Question. How many 4 digit numbers divisible by 5 using the digits 0 to 9 when:
(i) Repeated of digits not allowed
(ii) Repeated of digits allowed
Answer: Digits available: 0 to 9.
Required: 4-digit numbers divisible by 5.

1. When repetition of digits is not allowed:
For numbers divisible by 5, the units digit must be either 0 or 5.
- Case (i): Numbers ending with '0':
Units place is filled in 1 way. The thousands, hundreds, and tens places can be filled in \( 9 \), \( 8 \), and \( 7 \) ways respectively.
Number of ways \( = 9 \times 8 \times 7 \times 1 = 504 \).
- Case (ii): Numbers ending with '5':
Units place is filled in 1 way. The thousands place cannot be 0, so it can be filled in 8 ways. The hundreds and tens places can be filled in \( 8 \) and \( 7 \) ways respectively.
Number of ways \( = 8 \times 8 \times 7 \times 1 = 448 \).

\( \therefore \) Required 4-digit numbers divisible by 5 \( = 504 + 448 = 952 \) ans.

2. When repetition of digits is allowed:
- Units place can be filled in 2 ways ({0, 5}).
- Thousands place can be filled in 9 ways (excluding 0).
- Hundreds place can be filled in 10 ways.
- Tens place can be filled in 10 ways.
\( \dots \) Required 4-digit numbers divisible by 5 \( = 9 \times 10 \times 10 \times 2 = 1800 \) ans.

Question. How many numbers are there between 100 and 1000 such that every digit is either 2 or 9?
Answer: The numbers must be 3-digit numbers.
Digits available: 2 and 9.

Each of the three places (hundreds, tens, and units) can be filled in 2 ways (by either 2 or 9):
1. The hundreds place can be filled in 2 ways.
2. The tens place can be filled in 2 ways.
3. The units place can be filled in 2 ways.

\( \therefore \) The required 3-digit numbers \( = 2 \times 2 \times 2 = 8 \) ans.