Is My Approach to Solving This Permutation Problem Correct?

[bllnsr]

Homework Statement

http://i49.tinypic.com/wmmhbl.png

Homework Equations

The Attempt at a Solution

my answers :
Q1:(a)
(i) 8!
(ii) 4!*5!
Q1:(b)
7P3*5P2
Q2:
4320 ways
Q3:
12!*3!
Are my answers correct?

 

[Michael Redei]

Q1 looks fine. I haven't checked Q2, but Q3 is way off. How do you get that result anyway?

 

[haruspex]

Q1:(b)
7P3*5P2

P or C?

Q2:
4320 ways

Doesn't seem enough. How did you arrive at that?

Q3:
12!*3!

The mugs of each colour are identical, and you only care about the pattern of colours here.

 

[bllnsr]

Q2:
6!*3!=4320 ways

 

[haruspex]

Q2:
6!*3!=4320 ways

That doesn't count the ways one set of mugs may be placed in relation to the other set.

 

[bigQ]

In Q.1 (b) you should use combinations and not permutations since we are talking about selections. for Q.2 try to think placing china mugs in some specific no. of allowed spaces and then rearranging them. In Q.3 you can't consider all mugs different. Mugs of same colour can't be rearranged within themselves.

 

[bllnsr]

Ok so Q1(b) should be 7C3*5C2=350 ways right?

 

[haruspex]

Ok so Q1(b) should be 7C3*5C2=350 ways right?

Yes, as I suggested in post #3.

 

[bllnsr]

Q2:
There are 6 different plastic mugs and 3 different china mugs to be placed in a row
where no two china mugs are adjacent.

Place the 6 plastic mugs in row, inserting spaces before, after and between them.
_x_x_x_x_x_x_

There are 6! arrangements of the 6 plastic mugs.

Choose 3 of the 7 spaces and place the 3 china mugs.
There are 7P3 ways.

Therefore, there are:61*7P3 possible arrangements.

am i correct?

and as for Q3 I don't know how to start

 

[Michael Redei]

Q2 looks fine to me now.

Q3 contains a red herring, since first we're told that there are 3 identical red mugs, and then they should be regarded as one unit by being kept together. Imagine dumping the 3 red mugs in one large red box, then you have only 12 objects to arrange: the red box, 4 identical blue mugs and 7 identical yellow mugs.

 

[bllnsr]

@Michael Redei
Q3:
1 red box + 4 identical blue mugs + 7 identical yellow mugs = 12!
and there are 3 mugs in the red box so it should be 12!*3!

 

[Dick]

@Michael Redei
Q3:
1 red box + 4 identical blue mugs + 7 identical yellow mugs = 12!
and there are 3 mugs in the red box so it should be 12!*3!

Some of those mugs are identical! For example, there aren't 3! ways of arranging 3 identical red mugs. There's only one.

 

[haruspex]

Some of those mugs are identical! For example, there aren't 3! ways of arranging 3 identical red mugs. There's only one.

... furthermore, the question does not ask for the number of ways of arranging the mugs. It asks for the number of ways of arranging the colours.

 

[bllnsr]

Q3:
12!/7!*4!=3960 correct?

 

[haruspex]

Q3:
12!/7!*4!=3960 correct?

Yes!

 

[Michael Redei]

Q3:
12!/7!*4!=3960 correct?

It might look better as 12! / (7! 4!), or even better as $$\frac{12!}{7! \cdot 4! \cdot 1!} = 3960.$$
12! / 7! * 4! would be (12! / 7!) * 4! = 2 280 960

 

[bllnsr]

ok I got it and thanks for help everyone