You prepare decorations for a party. How many ways are there to arrange 12 blue balloons, 9 green lanterns and 6 red ribbons in a row, such that no two ribbons are next to each other?

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Mathematics

24 March, 02:34

You prepare decorations for a party. How many ways are there to arrange 12 blue balloons, 9 green lanterns and 6 red ribbons in a row, such that no two ribbons are next to each other?

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James Weber · Answer 1

Answer: 27! - [22! * 6!]

Step-by-step explanation:

Decorations available = 12 blue ballons, 9 Green Lanterns and 6 red ribbons

To determine the number if ways of arrangement for if the ribbons must not be together, we must first determine the number of possible ways to arrange these decorations items if there are no restrictions.

Total number of ways to arrange them = [12+9+6]! = 27! = 1.089 * 10^28

If this ribbons are to be arranged distinctively by making sure all six of them are together, then we arrange the 6 of them in different ways and consider the 6ribbons as one entity.

Number of ways to arrange 6 ribbons = 6!.

To arrange the total number of entity now that we have arranged this 6ribbons together and taking them as one entity become: = (12 + 9 + 1) !. = 22!

Number of ways to arrange if all of the ribbons are taken as 1 and we have 22 entities in total becomes: 22! * 6!.

Hence, to arrange these decoration items making sure no two ribbons are together becomes:

= 27! - (22! * 6!)

= 1.08880602 * 10^28