9 cards each of different colors include a pink card and a green card.
1.How many different arrangements do not have the pink card next to the green card?
it's easier to find the number that do have them next to each other. In that case, there are effectively 8 cards. 7 of the other colors, and one pink-green card.
So, there are 8! ways to arrange the "8" cards.
Then, you can swap the pink and green, so the total is 2*8!
Now just subtract that from the original 9! arrangements to get your answer.
To find the number of different arrangements that do not have the pink card next to the green card, we can use the concept of permutations.
First, let's calculate the total number of different arrangements of all 9 cards, without any restrictions. We can use the formula for permutations of distinct objects, which is n!, where n is the number of objects.
So, the total number of different arrangements of the 9 cards is 9!
To calculate the number of arrangements where the pink and green cards are together, we can consider them as one combined object. So, now we have 8 objects (pink and green as one) and find the arrangements, which will be 8!.
However, as we want to find the number of arrangements where the pink and green cards are not next to each other, we need to subtract the number from the total arrangements where they are together.
To calculate the number of arrangements where the pink and green cards are next to each other, we can consider them as one combined object, which can be arranged with the remaining 7 objects. So, the number of arrangements where the pink and green cards are together is 7!.
Finally, we subtract the arrangements where the pink and green cards are together from the total number of arrangements (without any restrictions) to get the number of arrangements where the pink and green cards are NOT next to each other.
Therefore, the number of different arrangements that do not have the pink card next to the green card is:
9! - 7!
You can now calculate this value using a calculator or by simplifying the equation further if needed.