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Question

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?

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Solution

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! ( nr )!r! .

Substitute 5 for n and 3 for r in the above formula.

C 5 3 = 5! ( 53 )!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( n1 )! n!=n( n1 )( n2 )![ n2 ]

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! ( nr )!r! .

Substitute 7 for n and 5 for r in the above formula.

C 7 5 = 7! ( 75 )!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( n1 )! n!=n( n1 )( n2 )![ n2 ]

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! ( nr )!r! .

Substitute 5 for n and 4 for r in the above formula.

C 5 4 = 5! ( 54 )!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! ( nr )!r! .

Substitute 7 for n and 4 for r in the above formula.

C 7 4 = 7! ( 74 )!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! ( nr )!r! .

Substitute 5 for n and 5 for r in the above formula.

C 5 5 = 5! ( 55 )!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! ( nr )!r! .

Substitute 7 for n and 3 for r in the above formula.

C 7 3 = 7! ( 73 )!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 .


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