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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 ?
Solution :
Given Variables & Initial Setup
Total number of students ($n$) = $25$.
Number of students to be chosen ($r$) = $10$.
Special group size ($k$) = $3$. These students have a condition: either all 3 join or none join.
This problem is partitioned into two mutually exclusive cases [Per the Addition Principle of Counting].
Step 1: Case I - All 3 students join the excursion
If all 3 specific students join the party, we have already filled $3$ spots out of the $10$ required spots.
Remaining spots to be filled = $10 - 3 = 7$.
Remaining pool of students available = $25 - 3 = 22$.
The number of ways to choose $7$ students from $22$ is given by the combination formula $_nC_r = \frac{n!}{r!(n-r)!}$.
Calculation: $_{22}C_7 = \frac{22!}{7!(22-7)!} = \frac{22!}{7! \times 15!}$
Substituting the values: $\frac{22 \times 21 \times 20 \times 19 \times 18 \times 17 \times 16}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} = 170,544$
Step 2: Case II - None of the 3 students join the excursion
If none of the 3 students join, we must fill all $10$ spots from the remaining pool of students.
Remaining pool of students = $25 - 3 = 22$.
Number of spots to be filled = $10$.
Calculation: $_{22}C_{10} = \frac{22!}{10!(22-10)!} = \frac{22!}{10! \times 12!}$
Substituting the values: $\frac{22 \times 21 \times 20 \times 19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13}{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} = 646,646$
Step 3: Calculating Total Combinations
According to the Fundamental Principle of Addition, since these two cases are mutually exclusive [Per the Addition Principle], we add the results of both cases:
$\text{Total ways} = (_{22}C_7) + (_{22}C_{10})$
$\text{Total ways} = 170,544 + 646,646$
$\text{Total ways} = 817,190$
Final Solution: The total number of ways the excursion party can be chosen is 817,190.
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 ?
- 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
Chapters in CBSE - Class 11 Mathematics
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