We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
232
2
avatar+85 

Henry's little brother has \(8\) identical stickers and \(4\) sheets of paper, each a different color. He puts all the stickers on the pieces of paper. How many ways are there for him to do this, if only the number of stickers on each sheet of paper matters? 

 

THANKS FOR ALL HELP

 

btw the answer is not 15

 Jan 20, 2019
 #1
avatar+5172 
+2

If the answer is not 15 then there must be some restriction on how many stickers must be on each piece of paper.

If papers are allowed to have 0-8 stickers then there are 15 ways to place them if only the number of stickers per page matters.

 

Maybe the pieces of paper are actually distinct, you just don't care about which stickers are on them.

In this case there will be \(\dbinom{11+3-1}{3-1} = \dbinom{11}{3}=165\) different arrangements.

 Jan 20, 2019
 #2
avatar+4249 
0

In fact, this is an application os stars and bars, so \(\binom{8+4-1}{3}=\binom{11}{3}=\boxed{165}.\)

.
 Jan 21, 2019

7 Online Users

avatar
avatar