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Question

(A) The number of ways to select at least one ball from $3$ identical red,$4$ identical blue,and $5$ identical green balls is given by the product of

Vedclass · Accepted Answer

$150$

Vedclass · Answer

(A) The number of ways to select at least one ball from $3$ identical red,$4$ identical blue,and $5$ identical green balls is given by the product of the number of choices for each color minus the case where no ball is selected: $x = (3+1)(4+1)(5+1) - 1 = 4 	imes 5 	imes 6 - 1 = 120 - 1 = 119$. For the examination,a student can either pass or fail in each of the $5$ subjects. The total number of outcomes is $2^5 = 32$. The student fails if they fail in at least one subject. The number of ways to fail is the total number of outcomes minus the case where the student passes all $5$ subjects: $y = 2^5 - 1 = 32 - 1 = 31$. Therefore,$x + y = 119 + 31 = 150$.

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