In an examination,Yamini has to select 4 ques | vedclass

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(C) To select $4$ questions from each part,we use the combination formula $C(n, r) = \frac{n!}{r!(n-r)!}$.Number of ways to select $4$ questions from

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$36750$

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(C) To select $4$ questions from each part,we use the combination formula $C(n, r) = \frac{n!}{r!(n-r)!}$.Number of ways to select $4$ questions from Part-$I$ ($6$ questions): $C(6, 4) = \frac{6!}{4!2!} = \frac{6 	imes 5}{2 	imes 1} = 15$.Number of ways to select $4$ questions from Part-$II$ ($7$ questions): $C(7, 4) = \frac{7!}{4!3!} = \frac{7 	imes 6 	imes 5}{3 	imes 2 	imes 1} = 35$.Number of ways to select $4$ questions from Part-$III$ ($8$ questions): $C(8, 4) = \frac{8!}{4!4!} = \frac{8 	imes 7 	imes 6 	imes 5}{4 	imes 3 	imes 2 	imes 1} = 70$.Total number of combinations = $C(6, 4) 	imes C(7, 4) 	imes C(8, 4) = 15 	imes 35 	imes 70 = 36750$.

In an examination,Yamini has to select $4$ questions from each part. There are $6, 7,$ and $8$ questions in Part-$I,$ Part-$II,$ and Part-$III,$ respectively. What is the number of possible combinations in which she can choose the questions?

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