Question

there are 6 multiple choise questions in an examination how many seqences of answers are possible if the first three questions have four choices each and the remaining three have three choices each by permuta tions and combination method?

Answer 1

Please see the attachment

Answer 2

Answer:

1728 ways

Step-by-step explanation:

There are 6 multiple choice questions (Given)

There are 4 choices for each of the first three questions (Given)

Therefore ,

The first 3 questions can be solved in 4 x 4 x 4 ways

i.e The first 3 questions can be solved in 64 ways

Also,

The other 3 remaining Questions have 3 choices each per question (Given)

Therefore ,

The remaining 3 questions can be solved in 3 x 3 x 3 ways

i.e The remaining 3 questions can be solved in 27 ways

Now , according to multiplication principle of counting ,

If there are m ways to do one task AND there are n ways to do another task , there are total ' m x n ' ways to do both tasks together .

Therefore ,

Total ways to solve all 6 questions containing 4 choices per question for the first 3 and 3 choices per question for the last 3 would be :

64 x 27 ways

= 1728 ways

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