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(C) The first $3$ questions each have $4$ choices. The number of ways to answer these is $4 	imes 4 	imes 4 = 4^3 = 64$.The next $3$ questions each

Vedclass · Accepted Answer

$8000$

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(C) The first $3$ questions each have $4$ choices. The number of ways to answer these is $4 	imes 4 	imes 4 = 4^3 = 64$.The next $3$ questions each have $5$ choices. The number of ways to answer these is $5 	imes 5 	imes 5 = 5^3 = 125$.According to the fundamental principle of counting,the total number of sequences of answers is the product of the number of ways to answer each set of questions.Total sequences = $64 	imes 125 = 8000$.

There are $6$ multiple choice questions in an examination. How many sequences of answers are possible,if the first three questions have $4$ choices each and the next three have $5$ choices each?

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