CBSE Class 11 Mathematics Permutations And Combinations Worksheet Set 09

CBSE Class 11 Mathematics Worksheet - Permutations and Combinations (8). 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. Consider the word "ASSASSINATION". Find:
1. Total number of words using all letters.
2. Number of words starting with A and ending with T.
3. Number of words starting with a vowel.
4. Number of words in which all vowels occur together.
5. Number of words in which all vowels never occur together.
6. Number of words in which all vowels occur together and all consonants occur together.
7. Number of words in which all 'S' are not together.
8. Number of words in which consonants occupy odd places.
9. Number of words such that there are always 4 letters between A and N.
10. If the different permutations of this word "ASSASSINATION" are listed in a dictionary, how many words are there before the first word starting with 'I'?
Answer: In the word "ASSASSINATION", total letters = 13 (Vowels: A = 3, I = 2, O = 1; Consonants: S = 4, N = 2, T = 1).

1. Total number of words using all letters:
\( = \frac{13!}{4!3!2!2!} = 10,810,800 \) ans.

2. Words starting with A and ending with T:
- Fix the position of 'A' in the first place and 'T' in the last place.
- The remaining 11 letters (containing S = 4, A = 2, N = 2, I = 2, O = 1) can be arranged in:
\( = \frac{11!}{4!2!2!2!} = 207,900 \) ans.

3. Words starting with a vowel:
There are 3 cases:
- Case I: Words starting with A:
\( = 1 \times \frac{12!}{4!2!2!2!} = 2,494,800 \)
- Case II: Words starting with O:
\( = 1 \times \frac{12!}{4!3!2!2!} = 831,600 \)
- Case III: Words starting with I:
\( = 1 \times \frac{12!}{4!3!2!} = 1,663,200 \)
\( \therefore \) Required number of words \( = 2,494,800 + 831,600 + 1,663,200 = 4,989,600 \) ans.

4. Words in which all vowels occur together:
- Group the 6 vowels (A, A, A, I, I, O) as 1 entity.
- We now arrange 8 entities (7 consonants + 1 block) in \( \frac{8!}{4!2!} = 840 \) ways.
- Within the block, the 6 vowels can be mutually arranged in \( \frac{6!}{3!2!} = 60 \) ways.
\( \therefore \) Required number of words \( = 840 \times 60 = 50,400 \) ans.

5. Words in which all vowels never occur together:
\( = \text{Total words} - \text{Words with vowels together} = 10,810,800 - 50,400 = 10,760,400 \) ans.

6. Words in which all vowels occur together and all consonants occur together:
- Group 6 vowels as 1 entity and 7 consonants as another entity. They can be arranged in \( 2! \) ways.
- Vowels can be mutually arranged in \( \frac{6!}{3!2!} = 60 \) ways.
- Consonants can be mutually arranged in \( \frac{7!}{4!2!} = 105 \) ways.
\( \therefore \) Required number of words \( = 60 \times 105 \times 2! = 12,600 \) ans.

7. Words in which all 'S' are not together:
- Group the four 'S' as 1 entity. We arrange 10 entities in \( \frac{10!}{3!2!2!} = 151,200 \) ways.
- Since the four 'S' are identical, they can be mutually arranged in only \( 1 \) way.
- Number of words with all 'S' together \( = 151,200 \times 1 = 151,200 \).
\( \therefore \) Required number of words \( = 10,810,800 - 151,200 = 10,659,600 \) ans.

8. Words in which consonants occupy odd places:
- There are 7 odd positions (1, 3, 5, 7, 9, 11, 13) for 7 consonants. They can be arranged in \( \frac{7!}{4!2!} = 105 \) ways.
- The 6 even positions are occupied by 6 vowels, which can be arranged in \( \frac{6!}{3!2!} = 60 \) ways.
\( \therefore \) Required number of words \( = 105 \times 60 = 6,300 \) ans.

9. Words such that there are always 4 letters between A and N:
- Placing 'A' in the first position and 'N' in the sixth position leaves 11 remaining places.
- The remaining 11 letters can be arranged in \( \frac{11!}{4!2!2!2!} = 207,900 \) ways.
- There are 8 such positional combinations. Thus, ways when A is first \( = 8 \times 207,900 = 1,663,200 \).
- Similarly, we have the same number of combinations when N comes before A.
\( \dots \) Required number of words \( = 1,663,200 + 1,663,200 = 3,326,400 \) ans.

10. Words before the first word starting with 'I' in a dictionary:
- In alphabetical order (A, I, N, O, S, T), the words before 'I' are those starting with 'A'.
- Fixing 'A' in the first position, the remaining 12 letters can be arranged in:
\( = 1 \times \frac{12!}{4!2!2!2!} = 2,494,800 \) ans.

Question. Word "MATHEMATICS" how many four letter words can be formed?
Answer: Total letters = 11 (M = 2, A = 2, T = 2, and H, E, I, C, S are 1 each). There are 8 distinct letters available.

There are 3 cases:
Case 1: All 4 letters are distinct:
- Select 4 distinct letters from 8 in \( ^8C_4 \) ways, and arrange them in \( 4! \) ways.
\( \therefore \text{Number of words} = ^8C_4 \times 4! = 70 \times 24 = 1680 \).

Case 2: Two of one kind and two different (e.g., 1 pair and 2 distinct letters):
- Select 1 pair out of 3 pairs (M, A, T) in \( ^3C_1 \) ways.
- Select 2 distinct letters out of the remaining 7 distinct letters in \( ^7C_2 \) ways.
- Arrange these 4 letters in \( \frac{4!}{2!} \) ways.
\( \therefore \text{Number of words} = ^3C_1 \times ^7C_2 \times \frac{4!}{2!} = 3 \times 21 \times 12 = 756 \).

Case 3: Two of one kind and two of another kind (2 pairs):
- Select 2 pairs out of 3 pairs in \( ^3C_2 \) ways.
- Arrange these 4 letters in \( \frac{4!}{2!2!} \) ways.
\( \therefore \text{Number of words} = ^3C_2 \times \frac{4!}{2!2!} = 3 \times 6 = 18 \).

\( \therefore \text{Total required words} = 1680 + 756 + 18 = 2454 \) ans.

Question. Word "ORDINATE". Find its rank?
Answer: 28988

Question. For the word "DAUGHTER", answer the following:
1. Total words using all letters.
2. Using 5 letters.
3. Start with A and end with E.
4. Start and end with consonants.
5. All vowels occur together.
6. All vowels never occur together.
7. All vowels together and all consonants occur together.
8. No two vowels are together.
9. Consonants occupy odd places.
10. There are always 2 letters between A and R.
11. Find the rank of word DAUGHTER.
12. ‘G & H’ are never together.
13. 5 letter words consisting of 2 vowels and 3 consonants.
Answer: [Questions as presented in textbook worksheet]

Question. For the word "INEFFECTIVE", answer the following:
1. Total words using all letters.
2. Start with ‘E’ and end with ‘T’.
3. All vowels occur together.
4. All vowels never together.
5. All vowels together and all consonants together.
6. No two vowels are together.
7. Start with a vowel.
8. These are always 3 letters between E & V.
9. All E not together.
10. (i) No. of words using only 4 letters (ii) Vowels occupy even places.
Answer: [Questions as presented in textbook worksheet]

Question. Word AGAIN. If all the letters of their word are arranged in a dictionary then what will be the 50th word?
Answer: Total letters = 5 (A = 2, G = 1, I = 1, N = 1). Alphabetical order of letters is: A, G, I, N.

- Number of words starting with A: \( 4! = 24 \)
- Number of words starting with G: \( \frac{4!}{2!} = 12 \)
- Number of words starting with I: \( \frac{4!}{2!} = 12 \)
Total words starting with A, G, and I \( = 24 + 12 + 12 = 48 \).

The words starting with N follow:
- 49th word: N, A, A, G, I
- 50th word: N, A, A, I, G
\( \therefore \) NAAIG is the 50th word.

Question. For a set of 5 true/false questions, no student has written all correct answers and no two students have given the same answers. What is the maximum number of students in the class?
Answer:
- A single true/false question can be answered in 2 ways.
- Therefore, there are 2 ways to answer each of the 5 questions.
- Total different answer sequences possible \( = 2 \times 2 \times 2 \times 2 \times 2 = 2^5 = 32 \).
- Out of these 32 sequences, only 1 sequence contains all correct answers.
- Since no student has written all correct answers, the maximum possible students (each with a unique incorrect sequence) \( = 32 - 1 = 31 \) ans.

Question. There are 6 periods in each working day of a school. In how many ways can one arrange 5 subjects such that each subject is allowed at least one period?
Answer:
- There are 6 periods and 5 subjects.
- Since each subject must be allowed at least one period, exactly 1 subject must repeat twice.
- The repeating subject can be selected from the 5 subjects in \( ^5C_1 \) ways.
- Once selected, the 6 periods (containing 1 subject twice and 4 others once) can be arranged in \( \frac{6!}{2!} \) ways.
\( \therefore \) Required number of ways \( = ^5C_1 \times \frac{6!}{2!} = 5 \times 360 = 1800 \) ans.

Question. How many numbers greater than a million can be formed with the digits 2, 3, 0, 3, 4, 2, 3?
Answer: The given 7 digits are: 2, 3, 0, 3, 4, 2, 3 (2 occurs twice, 3 occurs three times, 0 occurs once, 4 occurs once). To be greater than a million (1,000,000), the number must be a 7-digit number (so it cannot start with 0).

- Total permutations of these 7 digits \( = \frac{7!}{2!3!} = \frac{5040}{12} = 420 \).
- Number of arrangements starting with 0 \( = \frac{6!}{2!3!} = \frac{720}{12} = 60 \).
\( \therefore \) Total required numbers greater than a million \( = 420 - 60 = 360 \) ans.