# Thread: Permutations and Combinations

1. ## Permutations and Combinations

I'm having a hard time grasping just how to apply my knowledge of counting methods to certain questions in my textbook. Although everything (seems to) make sense, the wording and layout of questions for this topic has a way of jumbling my understanding.

As I looked through the review section of my textbook only every second or third question I read I knew how to approach, despite them being on par in terms of difficulty with the others (as per the layout of the book). If you could help me with these questions, and maybe help patch the hole in my understanding along the way, that would be great.

1. Seven people are to be seated in a row. Calculate the number of ways in which this can be done so that two particular persons, A and B, always have exactly one of the others between them.
I know that the total number of ways the 7 people can be seated is 7! = 5040, but where do I go from here?

2. Determine the number of possible permutations (from digits 1 through to 9) when the digits 1 and 2 are together but not necessarily in that order.
First, I calculated the total number of permutations to be 9! = 362880. I figured that if 1 and 2 are together, we can count that as 1 digit; "12" or "21". Therefore, we now have only 8 digits in the set. The total way the new combined digit can be arranged is 2! = 2. I don't know where to go from here. I tried 9!/2(8!) but got 42 (the answer is 48).

3. There are 10 chairs in a row:
a) In how many ways can three people be seated?
10 x 9 x 8 = 720.
b) In how many of these will the two end chairs be occupied?
c) In how many of these will the two end chairs be empty?

4. All possible three-digit numbers are formed from the odd digits {1, 3, 5, 7, 9},
a) How many such numbers are possible if each digit is used only once?
5 x 4 x 3 = 60
b) How many of the numbers from part a are lager than 350?
I know that in order to achieve that, the number must start with either a 3, 5, 7, or 9, but I don't know how to go about that mathematically. I tried removing 1 from the set, leaving me only 4 numbers, and then doing 4 x 3 x 2 = 24, but was wrong.

Thanks for your help, it is greatly appreciated.

2. ## Re: Permutations and Combinations

For #1: Choose whether A is to the left of B or to the right of B (2 possibilities). Then, choose the position of the person on the left (A*B****, *A*B***,**A*B**,***A*B*,****A*B) and the five choices for when B is on the left. So, there are 10 possible ways for A and B to be positioned. Regardless of how A and B are positioned, there are 5 seats left. There are 5! ways to seat the remaining people. So, there are $\displaystyle 2\cdot 5 \cdot 5!$ ways of seating all seven people.

For #2: Don't count permutations where the 1 and 2 are not together. What you said about combining the 1 and 2 is correct. First, choose the order of the 1 and 2. Like you said, there are 2 possibilities. Then, you are permuting 8 "things". So, the answer is $\displaystyle 2\cdot 8!$.

For #3: b) You have 3 people. In the left-most chair, put one of them. In the right-most chair, put one of them. Then, choose of the remaining eight chairs to put the last person. So, that is $\displaystyle 3\cdot 2\cdot 8$.
c) If the two end chairs are empty, then you only have 8 chairs. So, that is $\displaystyle 8\cdot 7\cdot 6$ ways to seat three people so that the end chairs are empty.

For #4: b) Count how many are larger than 400. That means the first digit must be 5, 7, or 9. So, the first digit has 3 choices. The second and third digits each have 5 choices. That is $\displaystyle 3\cdot 5\cdot 5 = 75$ 3-digit numbers larger than 400. Next, count how many are larger than 350 but less than 400. That means the first digit must be 3, the second digit may be 5, 7, or 9, and the last may be anything. That is $\displaystyle 1\cdot 3\cdot 5 = 15$ 3-digit numbers. All together, there are $\displaystyle 75+15 = 90$ 3-digit numbers larger than 350 using only odd digits. Note that this is not a permutation problem. Digits may be repeated.

Edit: For #4 a) Digits may be repeated. So, there are $\displaystyle 5^3 = 125$ 3-digit numbers using only odd digits.

3. ## Re: Permutations and Combinations Originally Posted by SlipEternal For #4: b) Count how many are larger than 400. That means the first digit must be 5, 7, or 9. So, the first digit has 3 choices. The second and third digits each have 5 choices. That is $\displaystyle 3\cdot 5\cdot 5 = 75$ 3-digit numbers larger than 400. Next, count how many are larger than 350 but less than 400. That means the first digit must be 3, the second digit may be 5, 7, or 9, and the last may be anything. That is $\displaystyle 1\cdot 3\cdot 5 = 15$ 3-digit numbers. All together, there are $\displaystyle 75+15 = 90$ 3-digit numbers larger than 350 using only odd digits. Note that this is not a permutation problem. Digits may be repeated.
What you've written makes sense, but my textbook says the answer is 36. However, it is very possible that the book is wrong.

4. ## Re: Permutations and Combinations Originally Posted by Fratricide What you've written makes sense, but my textbook says the answer is 36. However, it is very possible that the book is wrong.
If the question is asking how many 3-digit numbers have only odd digits and no repeated digits, then the answer would be:
Greater than 400: first digit may be 5,7,9, and other digits may be anything. After the first digit is chosen, there are 4 numbers left. After the second number is picked, there are 3 numbers left. So, that is $\displaystyle 3\cdot 4\cdot 3 = 36$.
Greater than 350, but less than 400: First digit must be 3, the second digit may be 5, 7, or 9. The last digit may be any of the remaining 3 digits. So, that is $\displaystyle 1\cdot 3\cdot 3 = 9$.

Total 3-digit numbers with no repeated digits and only odd digits that are greater than 350: $\displaystyle 36+9 = 45$.

The 36 from the book does show up, but I am guessing your book is wrong.

5. ## Re: Permutations and Combinations

Thanks for the help.

6. ## Re: Permutations and Combinations

Hello, Fratricide!

1. Seven people are to be seated in a row. Calculate the number of ways in which this can be done
so that two particular persons, A and B, always have exactly one of the others between them.

We have: $\displaystyle A\_\;\!B$ or $\displaystyle B\_\;\!A$ . . . 2 choices.

There are 5 choices for the "middle" person: $\displaystyle A\,\!X\!B$

Now we have 5 "people" to permute: .$\displaystyle \left\{\boxed{A\,\!X\!B},\, C,\,D,\,E,\,F\right\}$
There are 5! permutations.

Therefore, there are: .$\displaystyle 2\cdot5\cdot 5! \:=\:1200$ ways.

2. Determine the number of possible permutations (from digits 1 through to 9)
when the digits 1 and 2 are together, but not necessarily in that order.

Duct-tape the "1" and "2" together.
There are 2 possible orders: $\displaystyle 12$ or $\displaystyle 21.$

We have 8 "numbers" to permutate: .$\displaystyle \left\{\boxed{12},\,3,\,4,\,5,\,6,\,7,\,8,\,9 \right\}$
There are 8! ways.

Therefore, there are: .$\displaystyle 2\cdot8! \:=\:80,640$ permutations.

3. There are 10 chairs in a row:

a) In how many ways can three people be seated? . $\displaystyle {\color{blue}10\cdot9\cdot9 \,=\,720}$

b) In how many of these will the two end chairs be occupied?

The left chair can be occupied by any of the 3 people.
The right chair can be occupied by of the other 2 people.
The third person has a choice of the 8 middle chairs.

There are: .$\displaystyle 3\cdot2\cdot8 \,=\,48$ ways.

c) In how many of these will the two end chairs be empty?

The end chairs are to be empty.
The 3 people are to sit in the 8 other chairs.
There are: .$\displaystyle 8\cdot7\cdot6 \,=\,336$ ways.

4. All possible three-digit numbers are formed from the odd digits {1, 3, 5, 7, 9},

(a) How many such numbers are possible if each digit is used only once? .5 x 4 x 3 = 60

(b) How many of the numbers from part (a) are larger than 350?
Digits may NOT be repeated.

Suppose the number begins with "3".
The second digit must be 5, 7, or 9 . . . 3 choices.
The third digit can be any of the other 3 digits.
There are: .$\displaystyle 3cdot 3 \,=\,9$ numbers.

Suppose the number begins with 5, 7, or 9 . . . 3 choices.
The other two digits can be any of $\displaystyle 4\cdot3 \,=\,12$ choices.
There are: .$\displaystyle 3\cdot12 \,=\,36$ numbers.

Answer: .$\displaystyle 9 + 36 \,=\,45$ numbers.

combinations, permutations 