Plz help with counting
Miyu is giving out 9 identical chocolates to her 7 friends, including Dhruv. All possible distributions are equally likely. What is the probability that Dhruv gets exactly 2 chocolates?
+118680 ABCDEFG that is 7 friends
Cut into 9 peices that is 8 bars.
7+8=15
15C8=6435 ways to distribute.
If D gets 2 peices than you can take him and the 2 choc he is getting out
then you are left with
ABCEFG 6 friends divided into 7 parts that is 6 bars
6+6=12
12C6 = 924 possible ways
so that is 924/6435 = 28 / 195
I used the Stars and bars method to solve it.
I could have make a logic error. You need to decide for yourelf whether to accept this answer.
+2668 There are 9 stars (chocolate) and 7 - 1 = 6 bars.
This makes for \({15 \choose 6} = 5005\) cases.
If Dhruve gets exactly 2, there are now 7 stars, and 6 - 1 = 5 bars, which makes for \({12 \choose 5} = 792\) cases.
So, the probability is \({792 \over 5005} = \color{brown}\boxed{72 \over 455}\)
