1. ## Arrangements and combinations

1.In a true false test there are 12 questions. If a student decides to check six of each at random, in how many ways can she do it.

(I was thinking the answer would be 12 choose 6 or 12!/6!) Can anybody explain there reasoning in arriving to their answer.

2.in how many ways can a committee of 3 be chosen from 20 people? In how many ways can a president, a secretary and a treasurer be chosen?

(For the second part of this question i was thinking it was 20!, 19!, 18!)

Thanks again for any replies.

2. Originally Posted by multivariablecalc
1.In a true false test there are 12 questions. If a student decides to check six of each at random, in how many ways can she do it.
I do not understand what is meant by “a student decides to check six of each at random”
Does it mean “how many ways can she chose 6 questions to answer?”
Or does it mean “how many ways can she answer 6 questions?”
Or does it mean “how many ways can she chose and then answer 6 questions?”

Originally Posted by multivariablecalc
2.in how many ways can a committee of 3 be chosen from 20 people? In how many ways can a president, a secretary and a treasurer be chosen?
How many ways can a committee of 3 be chosen from 20 people: $\binom{20}{3}=\frac{20!}{3!\cdot 17!}$.

In how many ways can a president, a secretary and a treasurer be chosen? $\frac{20!}{17!}$

3. Hi Plato thanks for the reply. For the first problem i think the wording is really what has confused me for this question. I typed it verbatim out of the book. The Answer is (12 6) or a set of 12 choose 6.

For the second part of the second question why would be 17!? wouldn;t it be 20!19!18! becuase you have one person left to choose from? Thanks again for your response.

4. Originally Posted by multivariablecalc
For the second part of the second question why would be 17!? wouldn;t it be 20!19!18! becuase you have one person left to choose from? Thanks again for your response.
$20!=(20)(19)(18)(17!)$
So the answer is just $20\cdot 19\cdot 18$