There are six pairs of cups and saucers; two are red, two are white and two blue.
1. Ignoring the saucers, calculate the number of distinct arrangements of the cups.
2.Determine the number of distinct arrangements such that no cup is on a saucer of the same pattern for some choice of order for the saucers.
3. Calculate the probability that no cup is on a saucer of the same colour.
The Attempt at a Solution
(6!)/(2!2!2!)=90 possible arrangements
we need to look at all the possible scenarios
a) red cups are on the white saucers then the blue cups must be on the red saucers
b) Red cups are on the blue saucers then the white cups must be on the red saucers
c) If one red cup is on a white saucer and one red cup is on a blue saucer, then there are 2 choices for the white saucer and 2 choices for the blue saucer. The remaining white saucer must have a blue cup, and the remaining blue saucer must have a white cup; so there 2 ways to arrange the blue cup and white cup that are left on the red saucers.
This gives a total of 1+1+2*2*2=10 ARRANGEMENTS
Have i missed any arrangements?
10/90 = 1/9