I am working on some practice midterm problems. The solutions have been given, but I am a bit confused for a couple of them. I would really appreciate any help. Thanks in advance. Note: answers are in bold.

Question 1: This one I just had no clue, to me the number of last names could be limitless, so I am unsure how this conclusion can even be made?

Each person in a group of n people has a last name consisting of two uppercase letters. For what values of n can we guarantee that there are at least two people with the same last name?
(a) n≥26
(b) n≥52
(c) n ≥ 676
(d) n ≥ 677

Question 2: I had chosen b for this, as I thought to count the number of subsets of size 6 for a set S it would be ${n\choose 6}$?

What does the summation $\sum\limits_{k=7}^n {k-1\choose 6}$ count?
(a) Thenumber of subsets of {1, 2, . . . , n} having size 5.
(b) The number of subsets of {1, 2, . . . , n} having size 6.
(c) The number of subsets of {1, 2, . . . , n} having size 7.

Question 3: I understand here that the total number of answers (right and wrong) = $4^{17}$, but I don't understand why the chances of getting all the answers correct is only 1/$4^{17}$, wouldn't it be 17/$4^{17}$?

If you answer each question in this midterm by choosing an answer uniformly at random, what is the probability that you get all answers correct? (Note the midterm has 17 questions and each question has 4 options)
(a) 1/$17^{4}$
(b) 1/$4^{17}$
(c) $3^{17}$/$4^{17}$
(d) $4^{17}$/$3^{17}$


In question 1, there are only $26^2$ possible distinct last names consisting of two uppercase letters, and you need more than this to ensure duplicates.

In question 2, it turns out ${n\choose 7} = \sum\limits_{k=7}^n {k-1\choose 6}.$

In question 3, there is no reason to multiply by $17$. The probability of getting the first right is $1/4$, of getting the first two right is $1/4^2$ etc.


For question 3: For each question there are $4$ possible choices, therefore the probability of guessing a correct answer is $\frac14.$ Since there are $17$ questions, the probability of guessing all of them correct is $\left(\frac14\right)^{17}=\frac{1}{4^{17}}$


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