There are eight color cards with number, three blue cards blue1, blue1, blue1 and red cards red1, red1, red2, red2, red3.
I arrange them in a line.
Answer the following questions about the way of arrangement that there are just four places where the card of different color adjoins each other like red2-blue1-blue2-red1-red1-red3-blue1-red2 for example.
(1) When distinguishing the color of a card and not distinguishing a number, how many ways of arrangement is there?
(2) When distinguishing both the color of a card and a number, how many ways of arrangement is there?
I arrange them in a line.
Answer the following questions about the way of arrangement that there are just four places where the card of different color adjoins each other like red2-blue1-blue2-red1-red1-red3-blue1-red2 for example.
(1) When distinguishing the color of a card and not distinguishing a number, how many ways of arrangement is there?
(2) When distinguishing both the color of a card and a number, how many ways of arrangement is there?
Answer
Case 1 : A set of two blue and one blue are put in order like the example of problem sentence.
If a set of blue or one blue are put in A or F, there are only three places where the card of different color adjoins each other.
Thus, in this case, both ends of the arrangement should be reds.
Then ways of arranging the one blue and a set of two blue in the B ~ E are 4 × 3 = 12 ways.
Case 2 : Three blue cards are put in separately in A ~ F.
As for the way of arranging which suits the conditions in question, the both ends must be blue.
Since the remaining one blue card should be in either of B ~ E, it is four ways.
Therefore there are 12 + 4 = 16 ways in total.
There are five ways for red3 to be put on one of five places.
And there are （4 × 3） / (2 × 1) = six ways for two cards of red1 to be put on two of four remaining places.
Thus, 5 × 6 = 30 ways.
Since there is 30 ways of arranging red cards in each 16 ways as determined in (1), total ways of arrangement to be obtained as 16 × 30 = 480 ways.
(1) 16 ways
(2) 480 ways
(2) 480 ways
Solution
(1) It is considered that putting five red cards in order as shown in a figure and arranging blue cards in A~F.
Case 1 : A set of two blue and one blue are put in order like the example of problem sentence.
If a set of blue or one blue are put in A or F, there are only three places where the card of different color adjoins each other.
Thus, in this case, both ends of the arrangement should be reds.
Then ways of arranging the one blue and a set of two blue in the B ~ E are 4 × 3 = 12 ways.
Case 2 : Three blue cards are put in separately in A ~ F.
As for the way of arranging which suits the conditions in question, the both ends must be blue.
Since the remaining one blue card should be in either of B ~ E, it is four ways.
Therefore there are 12 + 4 = 16 ways in total.
(2) Since blue cards are all blue1, it examines how many ways of arrangement of red cards which are put in order by distinguishing a red number.
There are five ways for red3 to be put on one of five places.
And there are （4 × 3） / (2 × 1) = six ways for two cards of red1 to be put on two of four remaining places.
Thus, 5 × 6 = 30 ways.
Since there is 30 ways of arranging red cards in each 16 ways as determined in (1), total ways of arrangement to be obtained as 16 × 30 = 480 ways.
