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 help a student select the questions?
Given that a student needs to answer 8 questions out of 12 questions in which 5 from part I and 7 from part II.
It is also told that student needs to answer at least 3 questions from each part.
The possible cases are the following:
i. 3 from part I and 5 from part II
ii. 4 from part I and 4 from part II
iii. 5 from part I and 3 from part II
Let us assume the no. of ways of answering be N.
⇒ N = (no. of ways of answering 3 questions from part I and 5 from part II) + (no. of ways of answering 4 questions from part I and 4 from part II) + (no. of ways of answering 5 questions from part I and 3 from part II)
⇒ N = ((5C3) × (7C5)) + ((5C4) × (7C4)) + ((5C5) × (7C3))
We know that 
,
And also n! = (n)(n – 1)......2.1
⇒ 
⇒ 
⇒ 
⇒ N = (10 × 21) + (5 × 35) + (35)
⇒ N = 420
∴ The no. of ways of answering the question paper is 420.