CBSE Class 11 Mathematics Permutations And Combinations Worksheet Set 10

CBSE Class 11 Mathematics Worksheet - Permutations and Combinations (9). Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Question. How many numbers greater than 1000 but not greater than 4000 can be formed with the digits 0, 1, 2, 3, 4 if:
(i) Repetition of digits is allowed
(ii) Repetition of digits is not allowed
Answer: We need to find the number of 4-digit numbers in the range \( 1000 < \text{no.s} \le 4000 \) using the digits \( \{0, 1, 2, 3, 4\} \).

Case 1: When repetition of digits is allowed:
- The thousands place can be filled by 1, 2, or 3 (3 ways).
- The hundreds place can be filled by any of the 5 digits (5 ways).
- The tens place can be filled by any of the 5 digits (5 ways).
- The units place can be filled by any of the 5 digits (5 ways).
\( \therefore \) Total numbers formed \( = 3 \times 5 \times 5 \times 5 = 375 \).
Among these 375 numbers, the number 1000 is included (which is not greater than 1000), but 4000 is not included.
\( \therefore \) The required number of 4-digit numbers \( = 375 - 1 \text{ (for 1000)} + 1 \text{ (for 4000)} = 375 \).

Case 2: When repetition of digits is not allowed:
- The thousands place can be filled by 1, 2, or 3 (3 ways).
- Since repetition is not allowed, the remaining positions (hundreds, tens, units) can be filled in \( 4 \), \( 3 \), and \( 2 \) ways respectively.
\( \therefore \) Required numbers \( = 3 \times 4 \times 3 \times 2 = 72 \) ans.

Question. How many natural numbers less than 1000 can be formed from the digits 0, 1, 2, 3, 4, 5 when a digit may be repeated any number of times?
Answer: Given digits are \( \{0, 1, 2, 3, 4, 5\} \). We can form 1-digit, 2-digit, and 3-digit numbers less than 1000:

- 1-digit numbers: There are 5 such natural numbers (excluding 0) \( = 5 \).
- 2-digit numbers: The tens place can be filled in 5 ways (excluding 0), and the units place can be filled in 6 ways \( = 5 \times 6 = 30 \).
- 3-digit numbers: The hundreds place can be filled in 5 ways (excluding 0), the tens place in 6 ways, and the units place in 6 ways \( = 5 \times 6 \times 6 = 180 \).
\( \dots \) The total required numbers less than 1000 \( = 5 + 30 + 180 = 215 \) ans.

Question. How many numbers divisible by 5 and lying between 4000 and 5000 can be formed from the digits 4, 5, 6, 7 and 8?
Answer: 25

Question. How many numbers are there lying between 3000 to 5000 which are divisible by 2:
(i) when repetition of digits is not allowed
(ii) when repetition of digits is allowed
Answer:
(i) When repetition of digits is not allowed:
We have two main cases for the thousands place (since the number must lie between 3000 and 5000, the thousands place can only be 3 or 4):
- Case I: Thousands digit is 3 (odd):
The thousands place is filled in 1 way (by 3). The units place must be filled with an even digit: \( \{0, 2, 4, 6, 8\} \) (5 ways). The remaining hundreds and tens places can be filled in \( 8 \) and \( 7 \) ways respectively.
Number of ways \( = 1 \times 8 \times 7 \times 5 = 280 \).
- Case II: Thousands digit is 4 (even):
The thousands place is filled in 1 way (by 4). The units place must be filled with a remaining even digit: \( \{0, 2, 6, 8\} \) (4 ways). The hundreds and tens places can be filled in \( 8 \) and \( 7 \) ways respectively.
Number of ways \( = 1 \times 8 \times 7 \times 4 = 224 \).
\( \therefore \) Total required numbers \( = 280 + 224 = 504 \) ans.

(ii) When repetition of digits is allowed:
The thousands place can be filled by 3 or 4 (2 ways). The hundreds place can be filled in 10 ways, the tens place in 10 ways, and the units place can be filled by any of the 5 even digits \( \{0, 2, 4, 6, 8\} \) (5 ways).
Total ways \( = 2 \times 10 \times 10 \times 5 = 1000 \).
Since 3000 is included in these combinations (and we need numbers strictly greater than 3000), we subtract 1:
Required numbers \( = 1000 - 1 = 999 \) ans.

Question. Find the number of numbers greater than a million, that can be formed with the digits 1, 2, 0, 2, 4, 2, 4.
Answer: 360

Question. How many 4-digit even numbers can be formed using digits 0 to 9 when repetition of digits is allowed and when not allowed?
Answer:
(i) When repetition of digits is not allowed:
We have two cases based on the units place:
- Case I: Units digit is 0:
The units place is filled in 1 way. The thousands place can be filled by any of the remaining 9 digits (9 ways). The hundreds and tens places can be filled in \( 8 \) and \( 7 \) ways respectively.
Number of ways \( = 9 \times 8 \times 7 \times 1 = 504 \).
- Case II: Units digit is non-zero even (2, 4, 6, or 8):
The units place can be filled in 4 ways. The thousands place can be filled by any of the remaining digits except 0 (8 ways). The hundreds and tens places can be filled in \( 8 \) and \( 7 \) ways respectively.
Number of ways \( = 8 \times 8 \times 7 \times 4 = 1792 \).
\( \therefore \) Total even numbers \( = 504 + 1792 = 2296 \) ans.

(ii) When repetition of digits is allowed:
The thousands place can be filled in 9 ways (excluding 0), the hundreds place in 10 ways, the tens place in 10 ways, and the units place in 5 ways (by \( \{0, 2, 4, 6, 8\} \)).
\( \therefore \) Required even numbers \( = 9 \times 10 \times 10 \times 5 = 4500 \) ans.

Question. How many numbers between 400 and 1000 can be formed with the digits 0, 2, 3, 4, 5, 6 if no digit is repeated?
Answer: 60

Question. Given 4 flags of different colours, how many different signals can be generated using at least 2 flags?
Answer: We are given 4 flags of different colours. A signal can be generated using 2, 3, or 4 flags:

- Signals using 2 flags: \( 4 \times 3 = 12 \)
- Signals using 3 flags: \( 4 \times 3 \times 2 = 24 \)
- Signals using 4 flags: \( 4 \times 3 \times 2 \times 1 = 24 \)
\( \therefore \) Total number of signals using at least 2 flags \( = 12 + 24 + 24 = 60 \) ans.

Question. In how many ways can 3 prizes be distributed among 4 boys when:
(i) No boy gets more than one prize
(ii) A boy may get any number of prizes
(iii) No boy gets all the prizes
Answer:
(i) No boy gets more than one prize:
- The first prize can be given in 4 ways.
- The second prize can be given in 3 ways.
- The third prize can be given in 2 ways.
\( \therefore \) Number of ways \( = 4 \times 3 \times 2 = 24 \) ans.

(ii) A boy may get any number of prizes:
- Each of the 3 prizes can be given in 4 ways.
\( \therefore \) Required number of ways \( = 4 \times 4 \times 4 = 64 \) ans.

(iii) No boy gets all the prizes:
- Total number of ways to distribute the prizes \( = 64 \).
- Number of ways in which a single boy gets all the 3 prizes \( = 4 \) (since there are 4 boys).
\( \therefore \) Required number of ways \( = 64 - 4 = 60 \) ans.

Question. Consider the word "ORDINATE". Find:
1. Total number of words using all letters.
2. Number of words using 5 letters.
3. Number of words starting with R and ending with T.
4. Using 5 letters, the number of words starting with D and ending with E.
5. The number of words starting and ending with a vowel.
6. Number of words in which the letter 'D' is not included.
7. Number of words in which all vowels are together.
8. Number of words in which all vowels are never together.
9. Number of words in which all vowels are together and all consonants are together.
10. Number of words in which no two vowels are together.
11. Number of words such that the letter 'R' is always next to 'A'.
12. Number of words in which consonants occupy odd places.
13. Number of words such that vowels and consonants alternate.
14. Number of words such that the letters A and R are not together.
15. Number of words such that there are always 2 letters between A and R.
Answer: In the word "ORDINATE", total letters = 8 (Vowels = O, I, A, E = 4; Consonants = R, D, N, T = 4).

1. Total words using all letters: \( 8! = 40,320 \) ans.

2. Words using 5 letters: \( ^8P_5 = \frac{8!}{(8-5)!} = \frac{40320}{6} = 6720 \) ans.

3. Words starting with R and ending with T: Fix 'R' at the first position and 'T' at the last position. The remaining 6 letters can be arranged in \( 6! \) ways.
Required words \( = 1 \times 6! \times 1 = 720 \) ans.

4. Words of 5 letters starting with D and ending with E: Fix 'D' at the first position and 'E' at the last position. The remaining 3 positions must be filled from the remaining 6 letters.
Required words \( = 1 \times ^6P_3 \times 1 = 120 \) ans.

5. Words starting and ending with a vowel: There are 4 vowels. The first position can be filled in 4 ways, the last position in 3 ways, and the remaining 6 places can be arranged in \( 6! \) ways.
Required words \( = 4 \times 6! \times 3 = 8640 \) ans.

6. Words in which 'D' is not included: Excluding 'D', we have 7 letters remaining.
Required words \( = 7! = 5040 \) ans.

7. Words in which all vowels are together: Treat the 4 vowels (O, I, A, E) as a single letter. Now we arrange 5 entities (4 consonants + 1 block) in \( 5! \) ways. Within the block, the 4 vowels can be arranged in \( 4! \) ways.
Required words \( = 5! \times 4! = 120 \times 24 = 2880 \) ans.

8. Words in which all vowels are never together:
Required words \( = \text{Total words} - \text{Words with vowels together} = 40320 - 2880 = 37440 \) ans.

9. Words in which all vowels and all consonants are together: Arrange the group of vowels and consonants in \( 2! \) ways. Vowels can be mutually arranged in \( 4! \) ways, and consonants in \( 4! \) ways.
Required words \( = 2! \times 4! \times 4! = 1152 \) ans.

10. Words in which no two vowels are together: First, arrange the 4 consonants alternatively, which creates 5 available slots around them for vowels.
Required words \( = ^5P_4 \times 4! = 120 \times 24 = 2880 \) ans.

11. Words such that 'R' is always next to 'A': Treat "AR" as 1 non-interchangeable entity. We now arrange 7 entities (6 letters + 1 block) in \( 7! \) ways.
Required words \( = 1 \times 7! = 5040 \) ans.

12. Words in which consonants occupy odd places: The 4 consonants must be placed in the 4 odd positions (1, 3, 5, 7), which can be arranged in \( 4! \) ways. The 4 vowels must occupy the 4 even positions (2, 4, 6, 8), which can be arranged in \( 4! \) ways.
Required words \( = 4! \times 4! = 576 \) ans.

13. Words such that vowels and consonants alternate: There are two cases:
- Case I (V C V C V C V C): \( 4! \times 4! = 576 \) ways.
- Case II (C V C V C V C V): \( 4! \times 4! = 576 \) ways.
Required words \( = 576 + 576 = 1152 \) ans.

14. Words such that the letters A and R are not together:
Total words with A and R together \( = 7! \times 2! = 10080 \).
Required words \( = 8! - 10080 = 30240 \) ans.

15. Words such that there are always 2 letters between A and R: There are 5 positional pairs for A and R to have exactly 2 letters between them, with 2 possible orderings (A...R or R...A). The remaining 6 places can be filled in \( 6! \) ways.
Required words \( = 5 \times 2 \times 6! = 7200 \) ans.

Question. How many 5-letter words consisting of 3 vowels and 2 consonants can be formed?
Answer: We need to select 3 vowels out of 4, and 2 consonants out of 4, then arrange the selected 5 letters:
- Selection \( = ^4C_3 \times ^4C_2 \) ways.
- Arrangement of 5 selected letters \( = 5! \) ways.
\( \therefore \) Required words \( = ^4C_3 \times ^4C_2 \times 5! = 4 \times 6 \times 120 = 2880 \) ans.

Question. Find the RANK of the word INVOLUTE.
Answer: The alphabetical order of the letters in "INVOLUTE" is: E, I, L, N, O, T, U, V.
- Words starting with E: \( 7! = 5040 \)
- Words starting with I:
- Starting with I, E: \( 6! = 720 \)
- Starting with I, L: \( 6! = 720 \)
- Starting with I, N, E: \( 5! = 120 \)
- Starting with I, N, L: \( 5! = 120 \)
- Starting with I, N, O: \( 5! = 120 \)
- Starting with I, N, T: \( 5! = 120 \)
- Starting with I, N, U: \( 5! = 120 \)
- Starting with I, N, V, E: \( 4! = 24 \)
- Starting with I, N, V, L: \( 4! = 24 \)
- Starting with I, N, V, O, E: \( 3! = 6 \)
- Starting with I, N, V, O, L, E: \( 2! = 2 \)
- Starting with I, N, V, O, L, T: \( 2! = 2 \)
- Starting with I, N, V, O, L, U, E, T, V: 1
- Starting with "INVOLUTE" itself: 1

\( \therefore \text{Rank} = 5040 + 720 + 720 + 120 + 120 + 120 + 120 + 120 + 24 + 24 + 6 + 2 + 2 + 1 + 1 = 7140 \) ans.