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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 ?
Solution :
Given Variables & Initial Setup
The total number of questions is $12$, divided as follows:
- Part I ($P_1$): $5$ questions available.
- Part II ($P_2$): $7$ questions available.
The student must select a total of $8$ questions such that the condition "at least 3 from each part" is satisfied. [Per the Fundamental Counting Principle and Selection Principle].
Step 1: Identifying Possible Selection Scenarios
Since the total number of questions $n = 8$ and the requirement is $\ge 3$ from $P_1$ and $\ge 3$ from $P_2$, we evaluate the combinations of $(r_1, r_2)$ where $r_1$ is the number of questions from Part I and $r_2$ is the number of questions from Part II:
| Case | Questions from Part I ($r_1$) | Questions from Part II ($r_2$) | Total ($r_1 + r_2$) |
|---|---|---|---|
| Case 1 | $3$ | $5$ | $8$ |
| Case 2 | $4$ | $4$ | $8$ |
| Case 3 | $5$ | $3$ | $8$ |
Step 2: Calculating Combinations for Each Case
The number of ways to select $r$ items from $n$ is given by the combination formula $_nC_r = \frac{n!}{r!(n-r)!}$.
Calculation for Case 1 ($3$ from $P_1$, $5$ from $P_2$):
$_5C_3 \times _7C_5 = \frac{5!}{3!2!} \times \frac{7!}{5!2!} = 10 \times 21 = 210$Calculation for Case 2 ($4$ from $P_1$, $4$ from $P_2$):
$_5C_4 \times _7C_4 = \frac{5!}{4!1!} \times \frac{7!}{4!3!} = 5 \times 35 = 175$Calculation for Case 3 ($5$ from $P_1$, $3$ from $P_2$):
$_5C_5 \times _7C_3 = 1 \times \frac{7!}{3!4!} = 1 \times 35 = 35$Step 3: Total Summation
To find the total number of ways, we apply the Addition Principle of counting [Per the Rule of Sum in Combinatorics]:
Total Ways = $210 + 175 + 35$
Total Ways = $420$
Visual Representation of Selection Logic
Final Solution: The student can select the questions in 420 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 ?
- 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 ?
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