Probability (1 Viewer)

[acmilan]

In a library, there are 4 identical book A's, 5 identical book B's and 6 identical book C's in a row

i)how many arrangements are possible with no restriction
ii)how many arrangements are possible if all the same books are grouped together
iii)what is the probability that a copy of Book B and A will fill the end positions of the row

 

[kimmeh]

i) 15!/(4!5!6!)

ii) 3!
Since the books are identical, it doesnt matter how they are arranged within their own group. Its only the arrangement of groups that matters.

 

[gman03]

iii)13!/(3!4!6!) * 2 ?!?

 

[CM_Tutor]

Originally posted by gman03
iii)13!/(3!4!6!) * 2 ?!?

is the number of arrangements with A at one end, an B at the other.

So, the probability of this occuring if books arranged randomly is [13! * 2 / (3!4!6!)] / [15! / (4!5!6!)]
= [2 * 13! / (3!4!6!)] * (4!5!6! / 15!)
= 2 * (13! / 15!) * (5! / 3!) * 4!6! / (4!6!)
= 2 * [1 / (15 * 14)] * (4 * 5)
= 4 / 21

 

[~*HSC 4 life*~]

Originally posted by Asquithian
kimmeh do u know what 15 + 3 is
?

what are YOU doing here?

 

[gman03]

Originally posted by CM_Tutor
is the number of arrangements with A at one end, an B at the other.

Oh yeah, otherwise my probability will be greater than 1!!! That's a reason why I didn't do acturial studies.. probability.....

 

[kimmeh]

Originally posted by Asquithian
kimmeh do u know what 15 + 3 is
?

get out arts/law

 

[kimmeh]

Originally posted by Asquithian
dont be mean to me ....i want to be included

into maths? you rolled your eyes at me at USYD when you pulled out the uni maths/science books i had and said "i cant beleive your into this crap"