How to solve these two:

http://i49.tinypic.com/72a8ll.png

And is this question of combination?

http://i50.tinypic.com/ehdb9j.png

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- Dec 28th 2012, 12:23 AMbllnsrHow to solve this permutation question
How to solve these two:

http://i49.tinypic.com/72a8ll.png

And is this question of combination?

http://i50.tinypic.com/ehdb9j.png - Dec 28th 2012, 12:40 AMchiroRe: How to solve this permutation question
Hey bllnsr.

It's a lot easier to show us what you have tried with these kinds of problems because if the intuition is wrong, then an attempt will show how it is wrong.

Can you show us your attempts? - Dec 28th 2012, 12:54 AMbllnsrRe: How to solve this permutation question
Q2: it is written that china mugs are all separated from each other so I placed 6 plastic mugs and 3 china mugs alternately

6!*3!=4320 ways

Q3: This one confuses me and my answer to this question looks way off to me

12!*3!

and the last one I think is a question of combination :

7C3*5C3=350 ways - Dec 28th 2012, 06:02 AMSorobanRe: How to solve this permutation question
Hello, bllnsr!

I can't see Q2, but I'll guess what the problem says.

Quote:

Q2: it is written that china mugs are all separated from each other

so I placed 6 plastic mugs and 3 china mugs alternately: 6!*3!=4320 ways

There are 6plastic mugs and 3*different*china mugs to be placed in a row*different*

. . where no two china mugs are adjacent.

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

. . . . $\displaystyle \_\;x\;\_\;x\;\_\;x\;\_\;x\;\_\;x\;\_\;x\;\_$

There are $\displaystyle 6!$ arrangements of the 6 plastic mugs.

Choose 3 of the 7 spaces and place the 3 china mugs.

. . . There are $\displaystyle _7P_3$ ways.

Therefore, there are: .$\displaystyle (6!)(_7P_3) \:=\:151,\!200$ possible arrangements.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Suppose the 6 plastic mugs are identical and the 3 china mugs are, too.

Place the 6 plastic mugs in a row with the spaces.

Since they are identical, their order does not matter.

Now select 3 of the 7 spaces and place the 3 china mugs.

. . . There are $\displaystyle _7C_3$ ways.

Therefore, there are: .$\displaystyle _7C_3 \:=\:35$ possible arrangements.