Three identical yellow balloons,two identical red balloons and two identical blue balloons are strung in a row to celebrate Shema's birthday. Calculate the number of arrangements if:
The yellow balloons are next to each other and the blue balloons are not next to each other?
have a maths stats tmrw so am trying to just brushing up on a few concepts. would appreciate any help if possible :)
Hi, im new to web calc and just signed up, so please dont criticize me for this answer if its wrong :). (also I know very little latex so i will try to explain using words only)
First of all, the yellow balloons have to be together, so we can imagine all three as just one balloon. So, there are a total of $5!$ arrangements. But, we have forgotten that the blue balloons cant be together. There are $4$ cases in where the blue ballons are together, and we can also change the arrangements of the other $3$ ballons (because all $3$ yellow ballons are now one balloon), which is $3!$ arrangements, which is $6$, so we get a total of $24$ cases in which blue balloons are together, for a total of $96$ arrangements, which is the answer. The method which we have just used is called COMPLEMENTARY COUNTING. We calculated the opposite of what we want, and subtract it from the total amount.
Since the ballons of different colors are identical, we can't make arrangements inside of the $3$ yellow balloons. Lets say that the yellow balloons were distinguishable from one another, then we would get $3!$ more arrangements. But since that is not the case, $96$ arrangements is the answer.
(how did i do for my first answer? :D)