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The objective have to choose 3 tail results from the 8 events.
if we take O for T and 1 for H, then this can be imagined as an 8 -digit binary number composed of 3 zeros and 5 ones.
And we have to find that in how many different ways can we arrange 3 zeros in such a number?
To choose 3 from 8 we reckon like so:
There are 8 possibilities for the first "tail" ; 7 for the second and 6 for third. And, in the end we do not care about the order of these assignments. The calculation goes like so:
(8 * 7 * 6) /(3 * 2)
The more standard way to express this is:
8 ! /(5 ! * 3 !)=56
Alternatively, we can use the combination formula to choose 5 spots to place heads from 8 contiguous locations, that is given by
which gives the same answer.