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Q11:
In how many ways can the letters of the word ASSASSINATION be arranged so that all the $S$'s are together ?
Solution :
Step 1: Analyzing the Word Composition
The given word is ASSASSINATION. First, we identify the frequency of each distinct letter present in the word:
- A: 3
- S: 4
- I: 2
- N: 2
- T: 1
- O: 1
The total number of letters $n = 13$. [Based on the Fundamental Counting Principle for multiset permutations].
Step 2: Grouping the Constraints
The condition requires all $S$'s to be together. We treat the block of four $S$'s $(SSSS)$ as a single super-letter or entity, denoted as $X$.
The remaining letters are: $A, A, A, I, I, N, N, T, O$.
Total entities to arrange = $\{X, A, A, A, I, I, N, N, T, O\}$.
Number of entities = $1 + 3 + 2 + 2 + 1 + 1 = 10$.
Step 3: Calculating Permutations of the Entities
We must arrange these 10 entities, accounting for the repetitions of $A$ (3 times), $I$ (2 times), and $N$ (2 times). [Per the formula for permutations of a multiset: $\frac{n!}{n_1! n_2! ... n_k!}$].
The number of arrangements is given by:
$$P = \frac{10!}{3! \times 2! \times 2!}$$
Calculating the values:
$10! = 3,628,800$
$3! = 6$
$2! = 2$
$P = \frac{3,628,800}{6 \times 2 \times 2} = \frac{3,628,800}{24}$
$P = 151,200$
Step 4: Internal Arrangement of the S-Block
Since all four $S$'s are identical, the number of ways to arrange them within the block $X$ is $\frac{4!}{4!} = 1$. [By the logic of permutations of identical objects].
Step 5: Final Computation
The total number of arrangements is the product of the arrangement of the entities and the internal arrangements of the block.
$$\text{Total Ways} = 151,200 \times 1 = 151,200$$
Final Solution: 151,200 ways
More Questions from Class 11 Mathematics Permutations and Combinations Miscellaneous Exercise on Chapter 6
- Q1: How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER ?
- Q2: How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together?
- Q3(i): A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of: (i) exactly 3 girls ?
- Q3(ii): A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of: (ii) atleast 3 girls ?
- Q3(iii): A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of: (iii) atmost 3 girls ?
- Q4: If the different permutations of all the letter of the word EXAMINATION are listed as in a dictionary, how many words are there in this list before the first word starting with E ?
- Q5: How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated ?
- Q6: The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet ?
- Q7: In an examination, a question paper consists of 12 questions divided into two 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 ?
- Q8: Determine the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.
- Q9: It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible ?
- Q10: From a class of 25 students, 10 are to be chosen for an excursion party. There are 3 students who decide that either all of them will join or none of them will join. In how many ways can the excursion party be chosen ?
CBSE Solutions for Class 11 Mathematics Permutations and Combinations
Chapters in CBSE - Class 11 Mathematics
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