IIT JEE , AIEEE Forum - Mathematics , Algebra

nadeemoidu

I guess the answer given by akhil is right.

Here is a method which is not included in JEE but is very easy to understand. Both the methods give the same answer.

Let h(x) denote the no. of possible combinations with H at the end and t(x) denote the no. of possible combination with T at the end.

So the final answer will be h(13) + t(13)

Now, h(x) = t(x-1)    Because the only way to get a head at the end is to put H at the end of series ending with T.

Similarly t(x) = h(x-1 ) + t(x-1) because u can put a T at the end of any series.

so we can manually calculate the values upto 13

x    1   2 3 4 5 6     7    8
h(x) 1 1 2 3 5 8    13 21
t(x) 1 2 3 5 8 13 21 34

it is easy to see that this is the fibonacci series. h(13 ) + t ( 13 ) will be 610 .

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