[Discrete 2] Permutation/Combination Question

[MinusTheBear]

Homework Statement

I have two homework problems that seem similar, but I'm trying to understand another approach to the problem.

Question 1: One hundred tickets, numbered 1, 2, 3,..., 100, are sold to 100 different people for a drawing. Four different prizes are awarded, including a grand prize (a trip to Tahiti). How many ways are there to award the prizes if the people holding tickets 19 and 47 both win prizes?

Question 2: The English alphabet contains 21 consonants and five vowels. How many strings of six lowercase letters of the English alphabet contain exactly two vowels.

Homework Equations

The Attempt at a Solution

Attempt on solution 1: C(4,2) * 2 * 1 * 98 * 97
Attempt on solution 2: C(6,2) * 5^2 * 21^4

So, I got the same answer as the back of the book, but the back shows the solution for question 1 as C(4,1) * C(3,1) * 98 * 97.

I'm not sure I understand where they are getting the C(4,1), C(3,1) because using a similar approach on the second question doesn't work. IE C(6,1) * C(5,1) * 21^4. I see that the C(4,1) and C(3,1) is eliminating the positions available. But why doesn't this work similarly with question 2? Is it because the letters can be repeated?

Thanks.

 

[StoneTemplePython]

I'm not sure I understand where they are getting the C(4,1), C(3,1) because using a similar approach on the second question doesn't work. IE C(6,1) * C(5,1) * 21^4. I see that the C(4,1) and C(3,1) is eliminating the positions available. But why doesn't this work similarly with question 2? Is it because the letters can be repeated?

In a nutshell, I'd say yes, it's because of repetition issues.

You could think of problem 1 as being musical chairs, with a red, green, blue and yellow chair. 100 people play. You know person 19 and person 47 both are sitting when the music stops, how many possible arrangements are there? Notice how you 'use up' a chair each time someone sits down. Using the book approach: person 19 can be sitting in one of 4 chairs, then person 47 can sit in one of 3 remaining chairs, then you choose 2 from the remaining 98 people and allocate across 2, then one chairs. I.e. this gives $\Big(1*\binom{4}{1}\Big) * \Big(1*\binom{3}{1}\Big) * \Big(\binom{98}{2} * \binom{2}{1}*\binom{1}{1}\Big)=\binom{4}{1}*\binom{3}{1} * \Big(98 * 97\Big) = 4*3*98*97$.

I think problem 2 is a bit easier as it can be interpreted in terms of probabilities of coin tossing. Toss a biased coin with $\frac{5}{26}$ chance of heads and $\frac{21}{26}$ chance of tails. Now toss that same coin 6 times and record the results. (Notice we aren't "using up" anything here.) Any given sequence with 2 heads and 4 tails has probability of $\big(\frac{5}{26}\big)^2 \big(\frac{21}{26}\big)^4$ and there are $\binom{6}{2}$ ways to select such sequences in total, giving you $\Big(\binom{6}{2} \frac{5^2 21^4}{26^6}\Big)$ as total probability -- do you see how to convert this to the total number of outcomes you are estimating?

key idea: musical chairs and coin tossing are quited a bit different physically.

 

[PeroK]

They are two different problems. The first problem in particular can be solved several ways. I might look at it as:

19 can win any of four prizes; given that, 47 can win any of three prizes; given that, the next prize can be won by any of the 98; and, given that, the last prize can be won by any of the remaining 97. So:

$4 \times 3 \times 98 \times 97$

Which may be what the book intends. What do your numbers mean?