# [Discrete 2] Permutation/Combination Question

• MinusTheBear
In summary, the conversation discusses two homework questions that seem similar but require different approaches to solve. The first question involves awarding four prizes to 100 people, with two specific people winning. The answer can be calculated using combinations and accounting for the elimination of available positions. The second question involves creating strings of six lowercase letters with specific constraints. This can be solved using probability and the concept of coin tossing.
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.

## 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.

MinusTheBear said:
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.

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?

## 1. What is the difference between permutations and combinations?

Permutations are arrangements of objects where the order matters, while combinations are selections of objects where the order does not matter. For example, the permutations of the letters "ABC" are ABC, ACB, BAC, BCA, CAB, and CBA, while the combinations are ABC, ACB, BCA.

## 2. How do I calculate the number of permutations?

The number of permutations of n objects taken r at a time is given by nPr = n! / (n-r)!. For example, the number of permutations of 5 objects taken 3 at a time would be 5P3 = 5! / (5-3)! = 5! / 2! = 5*4*3 = 60.

## 3. How do I calculate the number of combinations?

The number of combinations of n objects taken r at a time is given by nCr = n! / (r!(n-r)!). For example, the number of combinations of 5 objects taken 3 at a time would be 5C3 = 5! / (3!(5-3)!) = 5! / (3!*2!) = 5*4*3 / (3*2) = 10.

## 4. Can permutations and combinations be used interchangeably?

No, permutations and combinations are different concepts and cannot be used interchangeably. Permutations refer to arrangements while combinations refer to selections, so they have different methods of calculation and cannot be used in the same situations.

## 5. How are permutations and combinations used in real life?

Permutations and combinations are used in a variety of fields, including mathematics, computer science, and statistics. In real life, they can be used to calculate the number of possible outcomes in a game of chance, the number of possible combinations of a lock, or the number of ways to arrange a set of data. They are also used in algorithms for tasks such as generating random numbers or finding the shortest route between two points.

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