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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 ?
Solution :
Step 1: Identifying the Frequency of Letters in 'EXAMINATION'
The word 'EXAMINATION' consists of 11 letters. We identify the frequency of each distinct letter as follows:
- Total letters ($n$) = 11
- E: 1
- X: 1
- A: 2
- M: 1
- I: 2
- N: 2
- T: 1
- O: 1
[By the Fundamental Principle of Counting and Permutations of Multisets, the total number of arrangements is given by $\frac{n!}{n_1! n_2! \dots n_k!}$].
Step 2: Determining the Alphabetical Order
To list the words in dictionary order, we arrange the unique letters of 'EXAMINATION' alphabetically:
A, E, I, M, N, O, T, X
Step 3: Calculating Words Before 'E'
A word in the dictionary starts with 'E' only after all words starting with 'A' have been exhausted. Therefore, we must calculate the total number of distinct arrangements that begin with the letter 'A'.
When the first position is fixed as 'A', we have 10 remaining positions to fill with the letters: {E, X, A, M, I, N, A, T, I, O, N}. Note: One 'A' is used, leaving us with:
- Remaining letters (10): E(1), X(1), A(1), M(1), I(2), N(2), T(1), O(1)
Step 4: Applying the Permutation Formula for Multiset
The number of ways to arrange the remaining 10 letters is given by the formula:
$$ \text{Arrangements} = \frac{10!}{n_E! \cdot n_X! \cdot n_A! \cdot n_M! \cdot n_I! \cdot n_N! \cdot n_T! \cdot n_O!} $$
Substituting the frequencies: $n_E=1, n_X=1, n_A=1, n_M=1, n_I=2, n_N=2, n_T=1, n_O=1$
$$ \text{Arrangements} = \frac{10!}{1! \cdot 1! \cdot 1! \cdot 1! \cdot 2! \cdot 2! \cdot 1! \cdot 1!} $$
$$ \text{Arrangements} = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 2} $$
$$ \text{Arrangements} = \frac{3,628,800}{4} = 907,200 $$
Step 5: Visualizing the Dictionary Placement
The number of words starting with 'A' represent all dictionary entries that appear strictly before the first entry starting with 'E'.
Final Solution: 907,200
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 ?
- 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 ?
- Q11: In how many ways can the letters of the word ASSASSINATION be arranged so that all the $S$'s are together ?
CBSE Solutions for Class 11 Mathematics Permutations and Combinations
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