Permutatuon question help PLEASE

[babacanoosh]

Permutatuon question help! PLEASE

1. How many ways can the word PAIRS be arranged if the first letter cannot be a vowel.



2. I believe you would just use counting, maybe (n!)/(n-r)!



3. 3x4x3x2x2=72. I put 3 as the first number because there are 3 different combinations not using a vowel. I believe this is right, but i don't understand why. Can somebody explain this to me? I came up with the rest of the numbers, knowing that 72 is the answer. But I'm not sure why i would use these combinations of numbers. Thanks


Second problem:

1. Using the letters from the word equation, how many five letter patterns can be formed in which q is followed immediately by u?

2. I believe you would just use counting, maybe (n!)/(n-r)!

3. I'm not even sure where to start with this one. I get thrown off by the 5 letter pattern even though there are 8 numbers. i tried many things and can't seem to come up with the right answer which is 580.

Thanks so much guys, I have a large test tomorrow and really appreciate the help.

 

[Dick]

There are, yes, 3 different ways to choose the first letter not a vowel. After that there are 4 ways to choose the second (since you've already used one letter), 3 for the third, etc. For the second, the pattern 'qu' can start in either the first, second, third or fourth position. Now fill in the other 3 letters. How many ways? Are you sure the answer isn't 480?

 

[babacanoosh]

yes you are right about 480, must have typed in the wrong number :/

ok so

would it be 2x2x6x5x4 = 480?

I don't understand exactly why you would have the number 2 for both possibilities if putting q and u in the front.

 

[Dick]

There are 4 places to put the 'qu'. So it's 4 times the number of ways to place the other three letters. How many ways to place the other three letters? 6 for the first, 5 for the second and 4 for the third. I'd write that as 4x6x5x4. Do you see why?

 

[babacanoosh]

yes i do now, i wasn't thinking as 'qu' as a unit taking up one space of the 5 spaces. Thanks a lot.

I have quite a few problems that I'm having difficulty on. Here is another. If i understand how to do this I may be able to do a lot of the others I cannot do.

1. 3 men and 3 women are to be seated in a row containing six chairs. How many seating arrangements are possible if the men and women are to sit in alternate seats?

2. counting or n!/(n!p!)

3. Correct answer=72
i tried counting using:
3x3x2x2x1x1=36.
I tried using the equation 6!/3!
=120.

Please help me out, thanks so much

 

[Dick]

Forget the equations for a while. So far your problems are easy enough that you hardly even need them. First, the men need three chairs and the women need three chairs. How many ways to choose those sets given the alternate chairs constraint?

 

[babacanoosh]

sorry I'm really not sure how to do this one.

the men wanting 3 chairs would be 6 different possibilities, and the women wanting 6 different possibilities would be 6 also. Given the alternate chairs restraint, would there be then 3 different combinations for men, and 3 different combinations for women?

 

[Dick]

Nope, there are two possibilities for the initial choice of chairs. The men can sit in the even numbered chairs and the women in the odd numbered chairs. Or vice versa. That's two possibilities. So start with '2x'. Now you have to multiply by the number of ways to seat the men in their chairs and the women in theirs.

 

[babacanoosh]

thanks, got it. Really appreciate it.

 

[Dick]

If you haven't already figured it out, these questions can be more of an art than a science, there's only a simple formula in the easiest of cases.

 

[Kushal]

you really should use logic.

you need to place arrange 3 men first. that's 3!

(you have to take the first man from a group of 3, which gives you 3 choices, then the second from a group of 2, giving you 2 choices, and then the last man from a group of 1 man, giving you only one possibility) you have then a total of 3 X 2 X 1 possibilities = 3!.

now you have to alternate using women. between every two men, you have to place a woman.

you can have one scenario: M W M W M W
between the first and second man, you have 3 possibilities (out of 3 women). then between the second and third, 2 possibilities. then just after the last man, one possibility (only one woman remains). the total number of ways = 3 X 2 X 1 = 3!


now, there is a second scenario: W M W M W M


the total possibilities is therefore 3! X 3! X 2.

3! = arranging the 3 men
3! = arranging the 3 women
2 = 2 possible scenarios.