Suppose to start with you have n distinguishable sticks (if it helps, visualize
ID: 3201569 • Letter: S
Question
Suppose to start with you have n distinguishable sticks (if it helps, visualize the sticks as having all different and unique colors, so there is green stick, blue stick etc). Each stick is broken into two parts, a long part and a short part (so now you have a long green part and short green part, a long blue part and a short blue part etc). These 2n parts are arranged into n pairs using which new sticks are formed. How many total outcomes are there in your sample space? That problem solved the case when n = 4. Suppose your event E was all long parts are paired with short parts. Thus here you would be ok if a blue long part got matched with a green short part. How many outcomes are there in your event E? A natural question at this point is why?!?!?! these combinatorics questions are tricky and make our brain hurt so why are we doing this? The problem above (as beautifully described in Feller's book on Probability [2]) comes from understanding genetic models in particular this paper [1], Here sticks correspond to chromosomes which when exposed to radiation correspond to breaking into parts where the longer part contains the centromere. If two "long" parts or two "short" parts combine then the cell dies.Explanation / Answer
a) There are 2n parts, out of which n parts are long and n are short. In the first case we are interested in forming pairs. It may happen here that long part get combined with long or short also. So, out of 2n parts, we have to select 2 parts and then make a pair of it. Therefore total number of ways in which we can do this is found out using combinations as 2nC2 =(2n)! / (2!*(2n-2)!), (which is equal to the total number of outcomes.)
b) Now, here in each pair, we want one short and one long part.
Long part can be found out in nC1 ways. imilarly short parts can be found in nC1 ways. Therefore total number of ways in which pair will have one short and one long part = (nC1)*(nC1) = n*n =n2 (which is equal to the total number of outcomes.)
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