Each of the 9 3 sets is equally likely to be drawn and 2 3 4 of them are successes so the probability of success is 2 3 4 9 3 24 84 2 7.
Three marbles with 2 colors can be aranged.
Fracisco s game involves 3 green 2 yellow 4 red and 3 black marbles.
All 12 marbles can be arranged permuted in 12.
So we raise three to the fourth power and subtract one to get eighty.
If he randomly draws three marbles from the bag without replacement what is the probability that he will draw yellow and then red and then black.
Many of these arrangements are identical because same colored marbles are indistinguishable from one another.
How many ways can i arrange 10 red marbles 5 white marbles and 6 blue marbles in a row.
The same 4 colors we ve picked them in different orders.
They can be numbers letters people colors etc.
In order to arrive at only those permutations that are distinct division is requi.
If the marbles are now unlabelled there are 4 possible labellings for the four white marbles 3 labellings for the three black marbles and 2 for the two red marbles.
1 slot 2 slot 3 slot and 4 slots.
Answer by edwin mccravy 18145 show source.
Example 15 in how many ways can 4 red 3 yellow and 2 green discs be arranged in a row if the discs of the same colour are indistinguishable.
So let s say we have 4 slots here.
Here there are three choices for which purple marble goes with the two blue ones and three choices for which blue marble goes with the two purple ones.
Now with that out of the way let s think about how many different ways we can pick 4 colors.
In how many different ways can these things be arranged in a row.
Since color are repeating so we use this formula 𝑛 𝑝1 𝑝2 𝑝3.
Any one of the three colors could be second.
A permutation of some number of objects means the collection of all possible arrangements of those objects.
Total number of discs 4 red 3yellow 2 green n 9.
This is the last question on my homework.
So each unlabelled order gives rise to 4 3 2 labelled orders.
Any one of the three colors could be third and any one of the three colors could be fourth unless the first three marbles are red in which case the fourth marble must be one of the other two colors.
To build such a set you could pick either of the 2 red marbles any one of the 3 white marbles and any one of the 4 blue marbles so there are 2 3 4 such sets.