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?
It is given that the question paper consists of 12 questions that are divided into two parts, where part I contains 5 questions and part II contains 7 questions.
Since a student is required to attempt 8 questions, selecting at least 3 from each part, thus the following arises:
(a) 3 questions from part I and 5 questions from part II to make total 8 questions.
(b) 4 questions from part I and 4 questions from part II to make total 8 questions.
(c) 5 questions from part I and 3 questions from part II to make total 8 questions.
In case (a), 3 questions are chosen from 5 questions in part I, thus the combination can be written as C 5 3 .
The formula for the combination is defined as,
C n r = n ! ( n − r ) ! r ! .
Substitute 5 for n and 3 for r in the above formula.
C 5 3 = 5 ! ( 5 − 3 ) ! 3 ! = 5 ! 2 ! 3 !
Cancel the common factors by factorizing the bigger term to the factorial.
The formula to calculate the factors of a factorial in terms of factorial itself is,
n ! = n ( n − 1 ) ! n ! = n ( n − 1 ) ( n − 2 ) ! [ n ≥ 2 ] ⋮
The combination can be written as,
C 5 3 = 5 × 4 × 3 ! 3 ! 2 × 1 = 5 × 4 2 × 1 = 10
Thus, the number of ways that the questions are chosen from part I is 10.
Also 5 questions are chosen from 7 questions in part II, thus the combination is C 7 5 .
The formula for the combination is defined as,
C n r = n ! ( n − r ) ! r ! .
Substitute 7 for n and 5 for r in the above formula.
C 7 5 = 7 ! ( 7 − 5 ) ! 5 ! = 7 ! 2 ! 5 !
Cancel the common factors by factorizing the bigger term to the factorial.
The formula to calculate the factors of a factorial in terms of factorial itself is,
n ! = n ( n − 1 ) ! n ! = n ( n − 1 ) ( n − 2 ) ! [ n ≥ 2 ] ⋮
The combination can be written as,
C 7 5 = 7 × 6 × 5 ! 5 ! 2 × 1 = 7 × 6 2 × 1 = 21
Thus, the number of ways that the questions are selected from part II is 21.
By multiplication principle which states that if an event can occur in m different ways and follows another event that can occur in n different ways, the number of ways that 3 questions are chosen from part I and 5 questions are chosen from part II is 10 × 21 = 210 .
In case (b), 4 questions are chosen from 5 questions in part I, thus the combination can be written as C 5 4 .
The formula for the combination is defined as,
C n r = n ! ( n − r ) ! r ! .
Substitute 5 for n and 4 for r in the above formula.
C 5 4 = 5 ! ( 5 − 4 ) ! 4 ! = 5 ! 1 ! 4 ! = 5 × 4 ! 4 ! = 5
Thus, the number of ways that the questions are chosen from part I is 5.
Also 4 questions are chosen from 7 questions in part II, thus the combination is C 7 4 .
The formula for the combination is defined as,
C n r = n ! ( n − r ) ! r ! .
Substitute 7 for n and 4 for r in the above formula.
C 7 4 = 7 ! ( 7 − 4 ) ! 4 ! = 7 ! 3 ! 4 ! = 7 × 6 × 5 × 4 ! 4 ! 3 × 2 × 1 = 35
Thus, the number of ways that the questions are selected from part II is 35.
By multiplication principle which states that if an event can occur in m different ways and follows another event that can occur in n different ways, the number of ways that 4 questions are chosen from part I and 4 questions are chosen from part II is 5 × 35 = 175 .
In case (c), 5 questions are chosen from 5 questions in part I, thus the combination can be written as C 5 5 .
The formula for the combination is defined as,
C n r = n ! ( n − r ) ! r ! .
Substitute 5 for n and 5 for r in the above formula.
C 5 5 = 5 ! ( 5 − 5 ) ! 5 ! = 5 ! 0 ! 5 ! = 1 1 = 1
Thus, the number of ways that the questions are chosen from part I is 1.
Also 3 questions are chosen from 7 questions in part II, thus the combination is C 7 3 .
The formula for the combination is defined as,
C n r = n ! ( n − r ) ! r ! .
Substitute 7 for n and 3 for r in the above formula.
C 7 3 = 7 ! ( 7 − 3 ) ! 3 ! = 7 ! 4 ! 3 ! = 7 × 6 × 5 × 4 ! 4 ! 3 × 2 × 1 = 35
Thus, the number of ways that the questions are selected from part II is 35.
By multiplication principle which states that if an event can occur in m different ways and follows another event that can occur in n different ways, the number of ways that 5 questions are chosen from part I and 3 questions are chosen from part II is 1 × 35 = 35 .
Thus the number of ways of selecting the questions from both parts is 210 + 175 + 35 = 420 .
It is given that the question paper consists of 12 questions that are divided into two parts, where part I contains 5 questions and part II contains 7 questions.
Since a student is required to attempt 8 questions, selecting at least 3 from each part, thus the following arises:
(a) 3 questions from part I and 5 questions from part II to make total 8 questions.
(b) 4 questions from part I and 4 questions from part II to make total 8 questions.
(c) 5 questions from part I and 3 questions from part II to make total 8 questions.
In case (a), 3 questions are chosen from 5 questions in part I, thus the combination can be written as
The formula for the combination is defined as,
Substitute 5 for n and 3 for r in the above formula.
Cancel the common factors by factorizing the bigger term to the factorial.
The formula to calculate the factors of a factorial in terms of factorial itself is,
The combination can be written as,
Thus, the number of ways that the questions are chosen from part I is 10.
Also 5 questions are chosen from 7 questions in part II, thus the combination is
The formula for the combination is defined as,
Substitute 7 for n and 5 for r in the above formula.
Cancel the common factors by factorizing the bigger term to the factorial.
The formula to calculate the factors of a factorial in terms of factorial itself is,
The combination can be written as,
Thus, the number of ways that the questions are selected from part II is 21.
By multiplication principle which states that if an event can occur in m different ways and follows another event that can occur in n different ways, the number of ways that 3 questions are chosen from part I and 5 questions are chosen from part II is
In case (b), 4 questions are chosen from 5 questions in part I, thus the combination can be written as
The formula for the combination is defined as,
Substitute 5 for n and 4 for r in the above formula.
Thus, the number of ways that the questions are chosen from part I is 5.
Also 4 questions are chosen from 7 questions in part II, thus the combination is
The formula for the combination is defined as,
Substitute 7 for n and 4 for r in the above formula.
Thus, the number of ways that the questions are selected from part II is 35.
By multiplication principle which states that if an event can occur in m different ways and follows another event that can occur in n different ways, the number of ways that 4 questions are chosen from part I and 4 questions are chosen from part II is
In case (c), 5 questions are chosen from 5 questions in part I, thus the combination can be written as
The formula for the combination is defined as,
Substitute 5 for n and 5 for r in the above formula.
Thus, the number of ways that the questions are chosen from part I is 1.
Also 3 questions are chosen from 7 questions in part II, thus the combination is
The formula for the combination is defined as,
Substitute 7 for n and 3 for r in the above formula.
Thus, the number of ways that the questions are selected from part II is 35.
By multiplication principle which states that if an event can occur in m different ways and follows another event that can occur in n different ways, the number of ways that 5 questions are chosen from part I and 3 questions are chosen from part II is
Thus the number of ways of selecting the questions from both parts is
