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There are 10 table tennis b***s consisting of 2 pink, 3 orange and 5 white ones, all the same size: 1. How many ways can they be arranged in a row. 2. In how many of these ways are the 2 pink ones separated from each other

Guest Jun 18, 2014

Best Answer 

 #8
avatar+26406 
+5

Yes, I think so.

Alan  Jun 18, 2014
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8+0 Answers

 #1
avatar+91471 
+5

This is what I think

1)   10!/(2!3!5!)

$${\frac{{\mathtt{10}}{!}}{\left({\mathtt{2}}{!}{\mathtt{\,\times\,}}{\mathtt{3}}{!}{\mathtt{\,\times\,}}{\mathtt{5}}{!}\right)}} = {\mathtt{2\,520}}$$

2)   I'm going to work out how many ways they are together

Just think of the 2 pin b***s as just 1 ball (they are glued together) but then multiply by 2 because they can be glued in 2 different ways

9!(3!5!)=  (This has been edited after Alan's comments)

$${\frac{{\mathtt{9}}{!}}{\left({\mathtt{3}}{!}{\mathtt{\,\times\,}}{\mathtt{5}}{!}\right)}} = {\mathtt{504}}$$

so the number of ways that they are not together is

$${\mathtt{2\,520}}{\mathtt{\,-\,}}{\mathtt{504}} = {\mathtt{2\,016}}$$

This is what I think but it may not be correct.

Melody  Jun 18, 2014
 #2
avatar+81051 
+5

Melody, I like your approach of treating the b***s like repeated letters in a word...that makes sense to me......but, we know what happened the last time we agreed on an answer (don't we??).....LOL!!!

 

CPhill  Jun 18, 2014
 #3
avatar+26406 
+5

Melody, in part 1 you didn't differentiate between the 2 pink b***s (.i.e. you treated pink1/pink2 as the same as pink2/pink1).  So why did you differentiate in part 2 (at least, it looks to me like you did, unless I've misinterpreted what you mean by being glued together in 2 different ways)?

Alan  Jun 18, 2014
 #4
avatar+91471 
0

Yes, it is always a worry Chris!

Melody  Jun 18, 2014
 #5
avatar+91471 
+5

I don't know Alan, probablility is one big mystery to me.

I thought I did. Isn't that what the 2! on the bottom is for? p1,p2=p2,p1 so I divided by 2?

Why don't you tell us what the answers are?

Melody  Jun 18, 2014
 #6
avatar+26406 
+5

Yes, in part 1 that's quite right.  But in part 2, you have pink1_glued_to_pink2 as different from  pink2_glued_to_pink1.  

 

Here's another way of thinking about part 2.

Imagine the two pink b***s are together in positions 1 and 2.  There are 8!/(3!*5!) = 56 ways of arranging the others.  Now move the pink b***s to positions 2 and 3.  There are another 56 ways of arranging the others.  Now move the pink b***s ...etc.  There are 9 positions where you can have the two pink b***s together in this way, so in total there are 9*56 = 504 ways the two pink b***s can be together (this is half your number).

All this assumes the positions are ordered of course.  For example it assumes that ppooowwwww is different from wwwwwooopp (you mustn't look at the b***s from the other side!!).

Alan  Jun 18, 2014
 #7
avatar+91471 
0

Okay - so I just have to drop the *2  then?

I think that makes sense. 

Melody  Jun 18, 2014
 #8
avatar+26406 
+5
Best Answer

Yes, I think so.

Alan  Jun 18, 2014

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