Question 274141
A tea set has 4 cups and saucers.Two of which are of same color and the other two are of different color.If the cups are placed randomly on the saucers.What is the probability that no cup is on a saucer of same color
<pre><font size = 4 color = "indigo"><b>
Suppose the cups are <font color = "red">A</font>, <font color = "red">B</font>, <font color = "blue">C</font>, and <font color = "blue">D</font>

Suppose the saucers are <font color = "blue">W</font>, <font color = "blue">X</font>, <font color = "red">Y</font>, and <font color = "red">Z</font>

where the color of the letter is the color of the cup or saucer,

Then there are only four ways to place them successfully

(<font color = "red">A</font>, <font color = "blue">W</font>), (<font color = "red">B</font>, <font color = "blue">X</font>), (<font color = "blue">C</font>, <font color = "red">Y</font>), (<font color = "blue">D</font>, <font color = "red">Z</font>)

(<font color = "red">A</font>, <font color = "blue">W</font>), (<font color = "red">B</font>, <font color = "blue">X</font>), (<font color = "blue">C</font>, <font color = "red">Z</font>), (<font color = "blue">D</font>, <font color = "red">Y</font>)

(<font color = "red">A</font>, <font color = "blue">X</font>), (<font color = "red">B</font>, <font color = "blue">W</font>), (<font color = "blue">C</font>, <font color = "red">Y</font>), (<font color = "blue">D</font>, <font color = "red">Z</font>)

(<font color = "red">A</font>, <font color = "blue">X</font>), (<font color = "red">B</font>, <font color = "blue">W</font>), (<font color = "blue">C</font>, <font color = "red">Z</font>), (<font color = "blue">D</font>, <font color = "red">Y</font>)

To place them any old way, there are 4 ways to place a cup on saucer <font color = "red">A</font>,
which leaves 3 ways to place a cup on saucer <font color = "red">B</font>, which leave 2 ways to place a
cup on saucer <font color = "blue">C</font>, which leaves 1 way to place a cup on saucer <font color = "blue">D</font>.
That's 4x3x2x1 = 4! = 24 ways

So the probability is 4 ways out of 24 ways or {{{4/24}}} or {{{1/6}}}

Edwin</pre>