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karthikeyan.srinivasan
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Hi there, karthikeyan.srinivasan! Beautiful problem (and the answer is correct).karthikeyan.srinivasan wrote:There are 21 identical T's and 19 identical E's. In howmany ways these items can be placed in a shelf such that no two E's are together?
Answer is 1540. But I dont know how?
Can anybody explain?
To help you understand my solution, let us do what I suggest to my students when they are dealing with big poisonous snakes (like this one): let us learn how to extract its poison from a "baby-snake" of the same species, right?!
Our baby-snake is: same problem, but with 5 T´s and 3 E´s...
You may start dealing with concrete possibities like the ones given bellow:
01) T T T E T E T E
02) T T E T T E T E
03) T E T T E T T E
etc... it´s hard NOT to miss anyone NOR to count the same twice, but if you get yourself organized (for instance start with "3 T´s together" then "2 groups of 2 T´s together" and finally "only 1 group of 2 T´s together") you will find the result for the baby-snake...
The fact is that we have to take out the baby-snake´s poison in a more "generalizable" way, that means it is NOT a good approach to COUNT EXPLICITLY its possibilities...it´s better to try to understand HOW the counting could be done without expliciting all scenarios!!
To do that, please consider the following squema:
x T x T x T x T x T x (where we have the 5 T´s and 6 x´s , where each x indicates a possibility to put one E there)!
From this squema, the 3 examples we have shown before are INCLUDED in the following sense:
01) x T x T x T E T E T E
02) x T x T E T x T E T E
03) x T E T x T E T x T E
Important: please note that x´s not substituted by E´s are simply "disconsidered" (because they are "potential places" not occupied)... in other words, you have to pay attention to how many T´s before the first, second and third E´s are placed, that´s all.
If you understand my squema and my examples, it´s not hard to understand that the answer for the baby-snake is C(6,3) = 20 because we are looking for how many ways we can put the 3 E´s between the 6 possible x´s ...
Now it´s really easy to find the answer to the big snake: C(22,19) = 1540. Think about it!
(If necessary, please feel free to ask a more-detail explanation in any part of my reasoning.)
Regards,
Fabio.
