How to solve these two:
http://i49.tinypic.com/72a8ll.png
And is this question of combination?
http://i50.tinypic.com/ehdb9j.png
How to solve these two:
http://i49.tinypic.com/72a8ll.png
And is this question of combination?
http://i50.tinypic.com/ehdb9j.png
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
Hello, bllnsr!
I can't see Q2, but I'll guess what the problem says.
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 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.
. . . . $\displaystyle \_\;x\;\_\;x\;\_\;x\;\_\;x\;\_\;x\;\_\;x\;\_$
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.