In how many ways can be the colours of the rainbow be arranged so that Red and Blue colours are always separated
Answers
Answer:sorry
Step-by-step explanation:I don't know heheheh you boy
Answer:
Answer is 3600.
The best way to answer a problem that deals with a restriction of two items not allowed together is to consider the complement. The complement of “blue and green not together” is “blue and green must be together”.
The number ways the different colours can be arranged such that blue and green must be together is 6!×2!=1440. This can be explained by thinking of the “blue and green being together” as one unit. Therefore, there will be 6 different units (i.e. colours) to be arranged in 6 positions, hence 6!. Also, the blue and green colours can be arranged in 2! ways within the unit.
Now, the total number of ways the 7 different colours can be arranged is 7!=5040.
Finally:
Numberofways“blueandgreencannotbetogether”=Totalnumberofwaysallthecolourscanbearranged−Numberofways“blueandgreenmustbetogether”=7!−6!⋅2!=5040−1440=3600.
421 views
As Dan Laver indicates, the best way to go about these problems is with the complement. However, with relatively small problems like this, you can do it straightforwardly if you really want.
First, place blue. If you put it first or last, then there is only one place that green can’t go, but if you put blue anywhere else, then there are spaces on both sides of blue, so green has two places it can’t go. This means you have to consider it in two cases:
Case 1: Blue is first or last (2 choices). This means green has 5 (7–blue-space next to blue) places it can go. Then we must place the remaining five colors: 5! arrangements. So Case 1 is: 2∗5∗5!=1200.
Case 2: Blue is anywhere except for first or last (5 choices). This means green has 4 (7-blue-both spaces next to blue) places it can go. Again, we must then place the remaining five colors. So Case 2 is: 5∗4∗5!=2400.
Add the two together: 2400+1200=3600