# Permutations are these not correct?

1. Oct 30, 2005

### rocketboy

Ok, for some reason my answers for the following 3 questions we were given for hw on permutations are wrong (according to the answers in the back of the text).

I was hoping that somebody could point me in the right direction, these basic ones should be simple for me. Thanks!

In how many ways can 11 players be seated on a bench so that Joey and Jill are not seated next to each other?

arrangements = total arrangements - arrangements where they are together
arrangements = 11! - 10!
arrangements = 36 288 000..........text had 32 659 200

In how many ways can four men and four women be seated around a circular table if each man must be flanked by two women?

i arranged the men first, then the women... i said that the first man has 8 possible seats, the second has 3 (because he can't sit beside another man and there has to be room for 2 women beside him) and the third has 2, and the last man has 1 possible seat. This leaves 4 seats for the first women, 3 for the second, 2 for the third, and 1 left for the last women to sit down.

thus, arrangements = (8 x 3 x 2 x 1) + (4!) = 72 text had 144

In how many ways can you form a three-digit number using only the digits of the number 21 150?

well, since it has to be 3 digits, 0 cannot be the first or second.
that leave's 4 digits to be put in the first and second places, and 5 for the last. divide by 2! because of the repeated 1.

thus, arrangements = (4 x 4 x 5) / 2 = 40 ways text has 26

-Jon

2. Oct 30, 2005

### QueenFisher

why can't the second number be 0? isn't 201 a 3-digit number?

3. Oct 30, 2005

### Muzza

11! - 2*10!. Joey can sit to the right or left of Jill.

4. Oct 30, 2005

### rocketboy

Thanks, I should have seen those.

But I still don't get the correct answer for the table question or the digits question.

if we're using the digits from 21 150 to form a 3 digit number....

then in the first spot we can have a 2, 1, or 5
in the second spot we can have a 2, 1, 5, or 0
and in the third spot we can also have a 2, 1, 5, 0

that means we can have 3 x 4 x 4 arrangements = 48 arrangements

but...there are repeats in that so we have to subtract those repeats which are:

if 2 appears in 3 or 2 spots = 3! spots
if 1 appears in 3 or 2 spots = 3! spots
if 5 appears in 3 or 2 spots = 3! spots
if 0 appears in 2 spots, that's 2 spots

so we subtract a total of 6+6+6+2 = 20???

i have to go for dinner...hehe...i will continue trying after.

Last edited: Oct 30, 2005
5. Oct 30, 2005

### S&S

1) Except Joey and Jill, you have 9 other players. Permute these 9 players. The 9 players have 10 possible space between them, then choose 2 out of these 10 space. Then permute Joey and Jill. Multiply all these possibilities:
9!2!(10C2)=as required.

2) For this kind circular sitting problem, you have to fix a people as a reference. I choose fix a man, then other men's position is 3 permute. Between these men you have 4 position for 4 women, then 4 permute:
3!4!= as required.

3) This one is tricky for me. It could be easy if the string we wanna for is same length as the original string. But we must form a 3 bit string from a 5 bit string in this question. I don't really know. Can someone help?

6. Oct 31, 2005

### S&S

Here is a computer generated possibility set.

101 125 215 520
102 150 250 521
105 151 251
110 152 501
112 201 502
115 205 510
120 210 511
121 211 512

7. Oct 31, 2005

### VietDao29

For #3, you can do it a little bit differently from what you did here. First, calculate the total ways to write a 3-digit number with only 2, 1, 5, 0.
That'll be:
3 * 3 * 2 = 18 ways.
This is the number of ways of writing a 3-digit number that has only 1 digit '1' in it.
Then calculate the total way that had 2 digits '1' in it.
* denotes some number chosen randomly from (2, 5, 0)
1*1 = 3 ways.
*11 = 2 ways.
11* = 3 ways.
Now add them together, and see what you get.
Is it clear enough?
Viet Dao,