+0

# PSL4#8

+1
290
4
+935

What I did:

Jul 28, 2019

#1
0

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Jul 28, 2019
edited by Guest  Jul 28, 2019
#2
+110167
+3

I am going to work with the idea that all the cookies are somehow different from each other and all the chips are as well

There are 4 places to put the first chip

4 for  the second

4 for the third and 4 for the forth.

That  is  4^4 = 256 possible combinations.

Now how many of those have one cookie with three and one cookie with one and 2 cookies with none?

There are 4 ways to chose 3 chips frm the four, ill melt those together so they cannot seperate.

Now there are 4 posible cookies to put that melted mess into.

Now there are 3 possible cookies left for the other one.

So that is 4*4*3 = 48 possibilies

How many ways can all 4 chips can go into one cookie.  That is easy .. 4 ways.

So there are 48+4 = 52 ways out of  256 ways that 3 or more chips are in any one cookie.

So   There must be  256-52 = 204 ways to distribute the chips so that there is NOT more than 2 chips in any one cookie.

ie       204/256 = 51/64

This answer is the same as your given one so I guess it is right.

NOTE: I had had not been trying to reproduce the given answer I most likely would have come up with some other solution.

To quote myself:

With probabillity it is easy to see why the correct answer is correct but often (it seems ) impossible to see why the wrong answer is wrong!  That is why it is so frustrating!

Jul 28, 2019
edited by Melody  Jul 28, 2019
#3
+110167
+2

I think that the reason 35 is wrong is that your method assumes that the 4 stars all all distinct and also assumes that the bars are not.

In this question neither are distinguishable

You can work it out like i have by assume all are distinct or you can probably do it some other way where neither are distinct but you can not go half way like you have done.

Jul 28, 2019
#4
+935
+2

I see. Thanks for your solution Melody!!!

dgfgrafgdfge111  Jul 28, 2019