There are 8 students appearing in an examinat | vedclass

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Question

(A) Total number of candidates $= 8$.First,arrange the $5$ candidates of different subjects in a row. They can be seated in $5!$ ways.$5! = 120$ ways.

Vedclass · Accepted Answer

$14400$

Vedclass · Answer

(A) Total number of candidates $= 8$.First,arrange the $5$ candidates of different subjects in a row. They can be seated in $5!$ ways.$5! = 120$ ways.These $5$ candidates create $6$ possible gaps (including the ends) where the $3$ Mathematics candidates can be seated so that no two Mathematics candidates sit next to each other.Number of ways to choose and arrange $3$ Mathematics candidates in these $6$ gaps is given by $P(6, 3)$.$P(6, 3) = \frac{6!}{(6-3)!} = 6 	imes 5 	imes 4 = 120$.By the fundamental principle of counting,the total number of ways is $5! 	imes P(6, 3) = 120 	imes 120 = 14400$.

There are $8$ students appearing in an examination of which $3$ have to appear in a Mathematics paper and the remaining $5$ in different subjects. In how many ways can they be made to sit in a row if the candidates in Mathematics cannot sit next to each other?

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